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2 years ago

PLEASE HELP!

Mathematics

1 answer:

melamori03 [73]2 years ago

4 0

Answer:

The answer is C.

Step-by-step explanation:

The $13 he started off with is the y-intercept or the initial value, which is the point where x=0.

The x-intercept, or rate of change, is $8.50.

The slope intercept formula is y=mx+b, where m=8.5 and b=13.

The equation for this problem is y=8.5x+13.

So the answer is C

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A carload of steel rods has arrived at Cybermatic Construction Company. The car contains 50,000 rods. Claude Ong, manager of Qua

amid [387]

Answer:

There is a 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch. This means that $\mu = 120, \sigma = 0.05$ .

Claude Ong, manager of Quality Assurance, directs his crew measure the lengths of 100 randomly selected rods. This means that $n = 100, s = \frac{\sigma}{\sqrt{n}} = \frac{0.05}{\sqrt{100}} = 0.005$ .

The probability that Claude's sample has a mean between 119.985 and 120.0125 inches is

We are working with the sample mean, so we use the standard deviation of the sample, that is, $s$ instead of $\sigma$ in the z score formula.

This probability is the pvalue of Z when $X = 120.0125$ subtracted by the pvalue of Z when $X = 119.985$ .

X = 120.0125

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{120.0125 - 120}{0.005}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938.

X = 119.985

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{119.985 - 120}{0.005}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0014.

So there is a 0.9938 - 0.0014 = 0.9924 = 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

7 0

2 years ago

A middle school band has 45 sixth and seventh graders The ratio of sixth to seventh graders is 2 to 3 how many sixth graders are

AnnyKZ [126]

Answer:

Step-by-step explanation:

There are 2+3 = 5 ratio units representing 45 band members, so each ratio unit represents 45/5 = 9 band members.

The 2 ratio units of sixth graders represent 2×9 = 18 sixth grade band members.

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2 years ago

Which system of equations can be used to find the roots of the equation 4 x Superscript 5 Baseline minus 12 x Superscript 4 Base

ehidna [41]

Answer:

D. on Edg 2020

Step-by-step explanation:

You are welcome, I helped better than that other guy!!!

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1 year ago

Body armor provides critical protection for law enforcement personnel, but it does affect balance and mobility. The article "Imp

yKpoI14uk [10]

Answer:

No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that true average task time with armor is less than 2 seconds.

Then, the null and alternative hypothesis are:

$H_0: \mu=2\\\\H_a:\mu< 2$

The significance level is 0.01.

The sample has a size n=52.

The sample mean is M=1.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.2}{\sqrt{52}}=0.028$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1.95-2}{0.028}=\dfrac{-0.05}{0.028}=-1.803$

The degrees of freedom for this sample size are:

$df=n-1=52-1=51$

This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

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1 year ago

The ratio of the lengths of the sides of △ABC is 3:6:7. M, N, and K are the midpoints of the sides. Perimeter of △MNK equals 7.4

artcher [175]

Answer:

AB=2.775

BC=5.55

CA=6.475

Step-by-step explanation:

Since midpoints split their sides in half, we can see that the triangle MNK formed by the midpoints will be half the perimeter of the triangle ABC. Since P of MNK = 7.4, we know that the perimeter of ABC = 7.4 * 2, which is 14.8. Now we can split the 14.8 so that it follows the ratio.

3+6+7=16

14.8/16=0.925

AB=0.925*3=2.775

BC=0.925*6=5.55

CA=0.925*7=6.475

8 0

2 years ago