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3 years ago

8

a bus travels 320 miles in 6.4 hours. if the bus continues at the same rate, which proportion can be used to find m, the number

of miles the bus will travel in 9 hours.

1 answer:

mr Goodwill [35]3 years ago

3 0

Answer:

A

Step-by-step explanation:

Ok first we need to find rate of MILES per HOUR

320/6.4=50

m= miles in 9 hours

9*50=450

<h3>Cross Multiply</h3>

is your answer

A

(6.4*9)/320m=Hours/miles

Hope this helps!

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Answer and explanation:

Given : Strands Copper wire from a manufacturer are analyzed forstrenghth and conductivity. The Results from 100 strands are as follows :

Strength Strength

High Low

High conductivity 74 8

Low conductivity 15 3

To find :

a) If a stand is randomly selected, the probability that is conductivity is high and its strength is high

The favorable outcome is 74

The probability is given by,

$P=\frac{74}{100}=0.74$

b) If a stand is randomly selected, the probability that its conductivity is low or strength is low

Conductivity is low A= 15+3=18

Strength is low B= 8+3=11

Conductivity is low and strength is low $A\cap B=3$

Probability is given by,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(A\cup B)=\frac{18}{100}+\frac{11}{100}-\frac{3}{100}$

$P(A\cup B)=0.18+0.11-0.03$

$P(A\cup B)=0.26$

c) Consider the event that a strand low conductivity and the event that the strand has a low strength. Are these tow events mutually exclusive?

Since the events the stand has low conductivity and the stand has low strength are not mutually exclusive, since there exists some cases in which both the events coincide. i.e. Intersection of both the events exists with probability 0.03.

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Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

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

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

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