MATHEMATICS COLLEGE

What is the answer too 4(2x+1)

Answers

Answer 1

Answer:

Answer:

8x+4

Step-by-step explanation:

4(2x+1)=8x+4

Answer 2

Answer:

Answer:

8x + 4

Step-by-step explanation:

4 x 2x = 8x

4 x 1 = 4

therefore its:

8x + 4

Brainliest is appreciated

Related Questions

Isaac works a job that pays him based oncommission. He earns 5% commission on allsales after the first $5000. To determine theamount of sales he earns commission on, heuses the function f(x) = x - 5000. Then he usesa different equation to determine how muchcommission he actually earns, g(x) = 0.05x. Hewants to create a composite that includesboth. What is the composite function?-=
In a study of plant safety, it was found that the time it took for machine operators to react to a warning light was normally distributed with a mean 1 second and standard deviation 0.3 second. Suppose a warning light visible to 4 operators goes on. What is the probability that the average (mean) reaction time of the 4 operators exceeds 1.25 seconds?
An angle is one-fourth of a circle. How many one-degree angles can be made from this angle?
A survey of cars on a certain stretch of highway during morning commute hours showed that 70% had only one occupant, 15% had 2, 10% had 3, 3% had 4, and 2% had 5. Let Xrepresent the number of occupants in a randomly chosen car.a. Find the probability mass function of X.b. Find P(X ≤ 2).c. Find P(X > 3).d. Find μX.e. Find σX
HELP ME PLZ THXxx u

A newspaper’s cover page is 3/8 text, and photographs fill the rest. If 2/
5 of the text is an article about
endangered species, what fraction of the cover page is the article about endangered species

Answers

Answer:

The answer is 0.15

Step-by-step explanation:

In this 2/5 of 3/8 is about endangered species

Therefore to obtain the answer we need to multiply the two fractions

2/5 × 3/8

Fraction of the cover page about endangered species is therefore = 2/5 × 3/8

= 0.15

Problem 9-10 The elongation of a steel bar under a particular tensile load may be assumed to be normally distributed, with a mean of .06 in. and standard deviation of .008 in. A sample of n=100 bars is subjected to the test. Find the probability that the sample mean elongation is between .0585 in. and .0605 in.

Answers

Answer:

The answer is 0.7036.

Step-by-step explanation:

Check the attached file for the computations.

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

What is z score?

Z score is used to determine by how many standard deviations the raw score is above or below the mean.

It is given by:

z = (raw score - mean) / (standard deviation÷√sample)

Mean = 0.06, standard deviation = 0.008, sample = 100.

For x = 0.0585:

z = (0.0585 - 0.06)/ (0.008 ÷√100) = -1.88

For x = 7.2:

z = (0.0605 - 0.06)/ (0.008 ÷√100) = 0.623

P(-1.88 < z < 0.63) = P(z < 0.63) - P(z < -1.88) = 0.7357 - 0.03 = 0.7057

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

Find out more on z score at: brainly.com/question/25638875

A clock is 17 minutes fast what is the correct time when it is showing 9:21 a.m.

Answers

Answer:

9:04

Step-by-step explanation:

21 - 17 = 4

it's using subtraction

Answer:

9:04

Step-by-step explanation:

9:21 - 17 minutes = 9:04

Use the distributive property to simplify this expression -2x(3x + x-5)

Answers

-6x-2x+10x
8x+10x
18x.

Answer:

−8x^2 + 10x

Step-by-step explanation:

It is believed that the average amount of money spent per U.S. household per week on food is about $98, with standard deviation $11. A random sample of 36 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average. State the null and alternative hypotheses for this test, the test statistic and determine if the results significant at the 5% level.

Answers

Answer:

The null hypothesis is

$H_o : \mu =$ $98

The alternative hypothesis is

$H_a : \mu >$ $98

test statistics $t_s = 1.091$

The the result of the test statistics is significant

Step-by-step explanation:

From the question we are told that

The population mean is $\mu =$ $98

The standard deviation is $\sigma =$ $11

The sample size is $n = 36$

The sample mean is $\= x =$ $100

The level of significance is $\alpha = 5$ % = 0.05

The null hypothesis is

$H_o : \mu =$ $98

The alternative hypothesis is

$H_a : \mu >$ $98

Now the critical values for this level of significance obtained from critical value for z-value table is $z_\alpha = 1.645$

The test statistics is mathematically evaluated as

$t_s = (\= x - \mu)/( (\sigma )/( √(n) ) )$

substituting values

$t_s = (100 - 98)/( (11 )/( √(36) ) )$

$t_s = 1.091$

Looking at $z_\alpha \ and \ t_s$ we see that $z_\alpha \ > t_s$ hence the we fail to reject the null hypothesis

hence there is no sufficient evidence to conclude that the mean weekly food budget for all households in this community is higher than the national average.

Thus the the result is significant

How many ways are there to add four positive odd numbers to get a sum of 22?

Answers

there are 18 ways to add four positive numbers to get 22

Other Questions

8/t + t^2; t=4 please help me

At Lincoln elementary school there are 88 students in grade 1 and 116 students in grade 2 grade 3 has 164 students is the number of students in grade 1 and grade 2 greater than the students in grade 3mn

B. 1/2 of 4/5 = ____ fifth(s)

Find the ratio of 4km to 1200m