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For what values of a does the equation ax^2+x+4=0 have only one real solution?

Answers

Answer 1

Answer:

Answer:

1/16

Step-by-step explanation:

To have one real solution, the discriminant must be 0.

b² − 4ac = 0

1² − 4a(4) = 0

1 − 16a = 0

a = 1/16

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Consider the following functions. f(x) = x − 3, g(x) = x2 Find (f + g)(x). Find the domain of (f + g)(x). (Enter your answer using interval notation.) Find (f − g)(x). Find the domain of (f − g)(x). (Enter your answer using interval notation.) Find (fg)(x). Find the domain of (fg)(x). (Enter your answer using interval notation.) Find f g (x). Find the domain of f g (x). (Enter your answer using interval notation.)

Answers

Answer:

$(f+g)(x)=x-3+x^2$ ; Domain = (-∞, ∞)

$(f-g)(x)=x-3-x^2$ ; Domain = (-∞, ∞)

$(fg)(x)=x^3-3x^2$ ; Domain = (-∞, ∞)

$((f)/(g))(x)=(x-3)/(x^2)$ ; Domain = (-∞,0)∪(0, ∞)

Step-by-step explanation:

The given functions are

$f(x)=x-3$

$g(x)=x^2$

$(f+g)(x)=f(x)+g(x)$

Substitute the values of the given functions.

$(f+g)(x)=(x-3)+x^2$

$(f+g)(x)=x-3+x^2$

The function $(f+g)(x)=x-3+x^2$ is a polynomial which is defined for all real values x.

Domain of (f+g)(x) = (-∞, ∞)

$(f-g)(x)=f(x)-g(x)$

Substitute the values of the given functions.

$(f-g)(x)=(x-3)-x^2$

$(f-g)(x)=x-3-x^2$

The function $(f-g)(x)=x-3-x^2$ is a polynomial which is defined for all real values x.

Domain of (f-g)(x) = (-∞, ∞)

$(fg)(x)=f(x)g(x)$

Substitute the values of the given functions.

$(fg)(x)=(x-3)x^2$

$(fg)(x)=x^3-3x^2$

The function $(fg)(x)=x^3-3x^2$ is a polynomial which is defined for all real values x.

Domain of (fg)(x) = (-∞, ∞)

$((f)/(g))(x)=(f(x))/(g(x))$

Substitute the values of the given functions.

$((f)/(g))(x)=(x-3)/(x^2)$

The function $((f)/(g))(x)=(x-3)/(x^2)$ is a rational function which is defined for all real values x except 0.

Domain of (f/g)(x) = (-∞,0)∪(0, ∞)

$(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

$(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

$(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

$f(g(x)) = x^2 - 3$ , domain: all real numbers.

To find (f + g)(x), we need to add the functions f(x) and g(x).

The function f(x) = x - 3 and the function $g(x) = x^2.$

So, $(f + g)(x) = f(x) + g(x) = (x - 3) + (x^2).$

Expanding this equation, we get $(f + g)(x) = x^2 + x - 3.$

To find the domain of (f + g)(x), we need to consider the domain of the individual functions f(x) and g(x).

Since both f(x) = x - 3 and $g(x) = x^2$ are defined for all real numbers, the domain of (f + g)(x) is also all real numbers.

To find (f - g)(x), we need to subtract the function g(x) from f(x).

So, $(f - g)(x) = f(x) - g(x) = (x - 3) - (x^2).$

Expanding this equation, we get $(f - g)(x) = -x^2 + x - 3.$

The domain of (f - g)(x) is also all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find (fg)(x), we need to multiply the functions f(x) and g(x).

So, $(fg)(x) = f(x) * g(x) = (x - 3) * (x^2).$

Expanding this equation, we get $(fg)(x) = x^3 - 3x^2.$

The domain of (fg)(x) is all real numbers, since both f(x) and g(x) are defined for all real numbers.

To find f(g(x)), we need to substitute g(x) into the function f(x).

So, $f(g(x)) = f(x^2) = x^2 - 3.$

The domain of f(g(x)) is also all real numbers, as $g(x) = x^2$ is defined for all real numbers, and f(x) = x - 3 is defined for all real numbers.

In summary:

- $(f + g)(x) = x^2 + x - 3$ , domain: all real numbers.

- $(f - g)(x) = -x^2 + x - 3$ , domain: all real numbers.

- $(fg)(x) = x^3 - 3x^2$ , domain: all real numbers.

- $f(g(x)) = x^2 - 3$ , domain: all real numbers.

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Can someone help me with this please?

Answers

let's see what to do...

Figure A is scaled image of Figure B.

So Figure A must be made in a proportion of Figure B.

We should find the ratio:

to find the ratio we have to use the same known sides.

$ratio = (5)/(2) \n$

Also we have :

$ratio = (12.5)/(x) \n$

So we have :

$(5)/(2) = (12.5)/(x) \n$

Multiply the sides by 2x :

$5 \: x = 25 \n x = 5$

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

g In a test of a gender-selection technique, results consisted of 200 baby girls and 14 baby boys. Based on this result, what is the probability of a girl born to a couple using this technique? Does it appear that the technique is effective in increasing the likelihood that a baby will be a girl? The probability that a girl will be born using this technique is approximately nothing. (Type an integer or decimal rounded to three decimal places as needed.) Does the technique appear effective in improving the likelihood of having a girl baby? No Yes

Answers

Answer:

- The probability of a girl born to a couple using this technique = 0.935

- Yes, the technique does appear effective in improving the likelihood of having a girl baby.

Step-by-step explanation:

1) In a test of gender selection, there are 200 girls and 14 baby boys.

To obtain the probability of a girl born to a couple using this technique

P(girls) = n(girls) ÷ n(total)

n(girls) = 200

n(total) = 200 + 14 = 214

P(girls) = (200/214) = 0.9346 = 0.935

2) Does it appear that the technique is effective in increasing the likelihood that a baby will be a girl?

We use an hypothesis test to confirm this. For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

Normally, the proportion of new girl babies and new boy babies should be close to each other (around 0.5 each), but this claim is that this gender selection technique favours the girl babies more than the male babies.

The null is that there is no significant evidence to conclude that the gender selection technique does favour more girl babies than boy babies.

The alternative hypothesis is that there is significant evidence to conclude that the gender selection technique does favour more girl babies than boy babies.

Mathematically,

The null hypothesis is represented as

H₀: p ≤ 0.50

The alternative hypothesis is given as

Hₐ: p > 0.50

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ)/σₓ

x = sample proportion = 0.935

μ = p₀ = The standard proportion we are comparing against = 0.50

σₓ = standard error = √[p(1-p)/n]

where n = Sample size = 214

p = 0.935

σₓ = √[0.935×0.065/214] = 0.0168521609 = 0.01685

t = (0.935 - 0.50) ÷ 0.01685

t = 25.81

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 214 - 1 = 213

Significance level = 0.05 (most tests are performed at this level)

The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for t = 25.81, at 0.05 significance level, df = 213, with a one tailed condition) = 0.000000001

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.000000001

0.000000001 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis & say that there is enough evidence to conclude that the gender selection technique does favour more girl babies than boy babies.

So, yes, the technique does appear effective in improving the likelihood of having a girl baby.

Hope this Helps!!!

What is the product of 4.201 and 5.3? Round to the nearest hundredth

Answers

Answer:

22.2653 or 22.27

Step-by-step explanation:

hope this helps

Answer: 22.265

Step-by-step explanation: 4.201x5.3 = 22.265

A bacteria culture is initially 10 grams at t=0 hours and grows at a rate proportional to its size. After an hour the bacteria culture weighs 11 grams. At what time will the bacteria have tripled in size?

Answers

A bacteria culture is initially 10 grams at t=0 hours & grows at a rate proportional to its size , After an hour the bacteria culture weighs 11 grams , The bacteria takes 11.56 hours to have tripled in size.

To find the time of bacteria when increasing the growth to tripled.

Given : when time=0 hours , weight=10 grams.

when time=1 hours , weight=11 grams.

To find: when time= ? hours , weight=30grams.

Here according to question, initial size = 10 grams we have asked for tripled in size i.e. 30 grams.

Now we knows that,

The formula for exponential growth in population or size is

$\rm (P)=P_0e^(rt)$ where,

$\rm P_0=initial\;size\n\nr= rate\;of\;growth\n\nt= time \;period$

Now, we put the value in formula we get,

$\rm P_0=10\;grams \n\nwhen ,\n\;\;t=1\;hour P(t)=11 grams\nThen,\n11=10e^{r(1)\n1.1 =e^r\n\n\rm Taking \;log(natural)\;both\;the\; side \;on \;solving\;we\;get,\nln(1.1)=r\;ln(e)\nr=ln(1.1)\nr=0.953101798043\approx0.095$

Now when the bacteria increase its size to triple

$\rm P(t) = 3 * 10 = 30$

Then, according to the formula we substitute values in the formula,

$\rm 30=10e^(0.095t)\n\n3=e^(0.095t)\n\nAgain \;we \;take\;natural\;log\;on \;both\;the\;sides, we\;get\nln\;3=0.095t\n\nt=(\rm ln\;3)/(0.095)\n\n\n\n\rm t= (1.09861228867)/(0.095) \n\n\ t=approx \; 11.56$

Therefore, The bacteria takes 11.56 hours to have tripled in size.

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Answer: It will take 11.56 hours .

Step-by-step explanation:

Exponential growth in population or size formula :

$P(t)=P_0e^(rt)$

, where $P_0$ = initial size

r= rate of growth

t= time period

As per given , we have

$P_0=10$ grams

At t= 1 , P(t)= 11 grams

Then,

$11=10e^(r(1))\n\n 1.1= e^r\n\n\text{Taking natural log on both sides , we get} \n\n\ln (1.1)=r\ln (e)\n\n r=\ln (1.1)\n\n r=0.0953101798043\approx0.095$

When, the bacteria have tripled in size , P(t) = 3 x10 = 30

Then,

$30=10e^(0.095t)\n\n 3=e^(0.095t)$

$\text{Taking natural log on both sides , we get}\n\n \ln 3=0.095t\n\n t=(\ln3)/(0.095)\n\n t=(1.09861228867)/(0.095)\approx11.56$

Hence, it will take 11.56 hours .

We wish to study the quality of our production line. We take a random sample of 1000 widgets from our line. A quality rating was determined for each of the 1000 widgets, and the average quality rating in the sample was 4. The sample standard deviation was 0.5. Also, 6 of the 1000 widgets were found to be defective.Estimate the average quality rating for widgets from our production line, and include the uncertainty in this estimate in the form, a ± b.

Answers

Answer:

Average quality rating was 4.54+-0.00549

Final answer:

The estimate for the average quality rating from the production line, given this sample, is 4, with a degree of uncertainty expressed by a 95% confidence interval of 4 ± 0.031. The confidence interval represents a range whereby we can be 95% confident that the true mean lies within.

Explanation:

Since you have the average (mean) quality rating and standard deviation from a sample size of 1000 widgets, we can use these statistics to establish an estimate for the entire production line. The estimate of the average quality rating is given as 4. However, to account for the uncertainty of our estimate due to it being based upon a sample rather than the entire population, we use the concept of a confidence interval.

The formula for a confidence interval is mean ± z* (standard deviation/sqrt(n)), where z is a z-score corresponding to our desired level of confidence. For simplicity, we can use a z-score of 1.96 to represent a confidence level of 95%.

Therefore, the uncertainty in this estimate (at 95% confidence) is calculated as:1.96 * (0.5/sqrt(1000)), approximately equal to 0.031. So the confidence interval for the average quality of widgets is 4 ± 0.031.

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