MATHEMATICS COLLEGE

1/2(x-6.5) = 3.16 what is x?

Answers

Answer 1

Answer:

Answer:

x =12.82

Step-by-step explanation:

1/2(x-6.5) = 3.16

Multiply each side by 2

1/2(x-6.5) *2 = 3.16 *2

(x-6.5) = 6.32

Add 6.5 to each side

(x-6.5+6.5) = 6.32+ 6.5

x =12.82

Answer 2

Answer:

Answer: 12.82

Step-by-step explanation:

1/2(x-6.5)= 3.16

x=12.82

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Determine whether lines L1 and L2 passing through the pairs of points are parallel, perpendicular, or neither. L1 : (–5, –5), (4, 6) L2 : (–9, 8), (–18, –3)

Write a simplified expression for the area of the rectangle below

Answers

Answer:

$= 12x + 40$

Step-by-step explanation:

$area = l * b \n = 20 * ((3)/(5) x + 2) \n = (60x)/(5) + 40 \n = 12x + 40$

hope this helps

brainliest appreciated

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HOT AIR BALLOON A balloon rises340 feet into the air. Then it descends

130 feet

PLEASE ANSWER I REALLY NEED IT

Answers

Answer:

it is at 210 feet

Step-by-step explanation:

it rises 340 feet, then it descends 130 so we subtract 130 from 340 which gets us 210 feet.

Suppose we express the amount of land under cultivation as the product of four factors:Land = (land/food) x (food/kcal) x (kcal/person) x (population)

The annual growth rates for each factor are:
1. the land required to grow a unit of food, -1% (due to greater productivity per unit of land)
2. the amount of food grown per calorie of food eaten by a human, +0.5%
3. per capita calorie consumption, +0.1%
4. the size of the population, +1.5%.

Required:
At these rates, how long would it take to double the amount of cultivated land needed? At that time, how much less land would be required to grow a unit of food?

Answers

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following annual growth rates:

land/food = - 1%

food/kcal = 0.5%

kcal/person = 0.1%

population = 1.5%

Σ annual growth rates = (-1 + 0.5 + 0.1 + 1.5)% = 1.1% = 0.011

Exponential growth in Land :

L = Lo * e^(rt)

Where Lo = Initial ; L = increase after t years ; r = growth rate

Time for amount of cultivated land to double

L = 2 * initial

L = 2Lo

2Lo = Lo * e^(rt)

2 = e^(0.011t)

Take the In of both sides

In(2) = 0.011t

0.6931471 = 0.011t

t = 0.6931471 / 0.011

t = 63.01 years

Land per unit of food at t = 63.01 years

L = Fo * e^(rt)

r = growth rate of land required to grow a unit of food = 1% = 0.01

L/Fo = e^(-0.01* 63.01)

L/Fo = e^(−0.6301)

= 0.5325385 = 0.53253 * 100% = 53.25%

Land per unit now takes (100% - 53.25%) = 46.75%

zoe has earned 650$ during the four weeks she worked at the rec center. the first 2 weeks she earned 220$ and 98$. the last 2 weeks she earned the same amount. how much money did zoe earn in the last 2 weeks

Answers

Answer:

The wording of this question was really confusing, but I believe the answer is 166 dollars.

Step-by-step explanation:

$220+ $98= $318

650-318=332

332/2=166

The brain volumes (cm cubedcm3) of 20 brains have a mean of 1189.81189.8 cm cubedcm3 and a standard deviation of 126.5126.5 cm cubedcm3. Use the given standard deviation and the range rule of thumb to identify the limits separating values that are significantly low or significantly high. For such data, would a brain volume of 1432.81432.8 cm cubedcm3 be significantly high?

Answers

Answer:

Lower limit = 936.8

Upper limit = 1442.8

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 1189.8 cube cm

Standard Deviation, σ = 126.5 cube cm

Range rule of thumb:

The range rule of thumb says that the range is four times the standard deviation.

$\text{Range} = 4* 126.5 = 506$

The range rule of thumb suggests that most values would be in the area covered by four standard deviations that is within two standard deviations above or below the mean.

Lower limit =

$\mu - 2\sigma\n= 1189.8 - 2(126.5)\n = 936.8$

Upper limit =

$\mu + 2\sigma\n= 1189.8 + 2(126.5)\n = 1442.8$

Thus, most values lie within (936.8,1442.8)

A brain volume of 1432.8 cube cm is not significantly high because it is less than the upper limit.

A study published in 1993 found that babies born at different times of the year may develop the ability to crawl at different ages! The author of the study suggested that these differences may be related to the temperature at the time the infant is 6 months old. (Benson and Janette, Infant Behavior and Development [1993]. The study found that 32 babies born in January crawled at an average age of 29.84 weeks, with a standard deviation of 7.08 weeks. Among 21 July babies, crawling ages averaged 33.64 weeks, with a standard deviation of 6.91 weeks. Is this difference significant?

Answers

Answer:

$t=\frac{(29.84-33.64)-0}{\sqrt{(7.08^2)/(32)+(6.91^2)/(21)}}}=-1.939$

$p_v =2*P(t_(51)<-1.939)=0.0580$

Comparing the p value with the significance assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different at 1% of significance.

Step-by-step explanation:

Data given

$\bar X_(1)=29.84$ represent the mean for sample January

$\bar X_(2)=33.64$ represent the mean for sample July

$s_(1)=7.08$ represent the sample standard deviation for 1

$s_(2)=6.91$ represent the sample standard deviation for 2

$n_(1)=32$ sample size for the group 2

$n_(2)=21$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_(1)-\mu_(2)=0$

Alternative hypothesis: $\mu_(1) - \mu_(2)\neq 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_(1)-\bar X_(2))-\Delta}{\sqrt{(\sigma^2_(1))/(n_(1))+(\sigma^2_(2))/(n_(2))}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=32+21-2=51$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(29.84-33.64)-0}{\sqrt{(7.08^2)/(32)+(6.91^2)/(21)}}}=-1.939$

P value

Since is a bilateral test the p value would be:

$p_v =2*P(t_(51)<-1.939)=0.0580$

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