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3 years ago

How might the lack of reliable transportation for getting to a job affect your decision to take it or not

1 answer:

8 0

Without reliable transportation, you could be late a lot, and get fired. That shows up when you apply for a new job, even if you get better transportation. If you don't want to get fired, you may not want to take the job.

You might be interested in

How many

MariettaO [177]

Answer:

Step-by-step explanation:

92/17= 5.4

there are 5 total 17s in 92.

6 0

3 years ago

The Davis family drove a total of 587 miles, starting on Friday and ending on Sunday. They drove 182 miles on Friday and 260 mil

valina [46]

Answer:

145

Step-by-step explanation:

Because what you want to do is add 260 to 182 then subtract from 587. I hope that helped! :)

8 0

2 years ago

Which is a perfect square? 61 62 63 65

Morgarella [4.7K]

Answer:

None

Step-by-step explanation:

None of these, but $\sqrt{64}=8$ so if 64 is an option than that's the correct option

$\sqrt{61} =7.81 \\\sqrt{62}=7.87\\\sqrt{63} =7.93\\\sqrt{65} =8.06$

3 0

3 years ago

suppose that you have 3000$ to invest. which investment yields the greater return over a 10 year period: 8.04% compounded daily

Ksenya-84 [330]

Answer: Option A: 8.04% compounded daily

Step-by-step explanation:

$A = P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\qquad where\\\\\bullet A = \text{accrued amount (principal plus interest earned)}\\\bullet P = \text{principal (amount invested)}\\\bullet r = \text{rate (in decimal form)}\\\bullet n=\text{number of times compounded in one year}\\\bullet t=\text{time (number of years)}\\\\\\Option\ A:\\A = \text{unknown}\\P=3000\\r=8.04\%\rightarrow 0.0804\\n=\text{daily}\rightarrow 365\\t=10\\\\A=3000\bigg(1+\dfrac{0.0804}{365}\bigg)^{365\times 10}\\\\.\ =\$ 6,702.79$

$Option\ B:\\A = \text{unknown}\\P=3000\\r=8.1\%\rightarrow 0.081\\n=\text{quarterly}\rightarrow 4\\t=10\\\\A=3000\bigg(1+\dfrac{0.081}{4}\bigg)^{4\times 10}\\\\.\ =\$ 6,689.37$

Option A results in the greater amount of money.

7 0

3 years ago

Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b

sesenic [268]

Answer:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

-The sample is too small to make judgments about skewness or symmetry.

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

$p_v =2*P(t_{(14)}$

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

Step-by-step explanation:

First we need to find the difference defined as:

(Operator 1 minus Operator 2)

d1=1.326-1.323=0.003 d2=1.337-1.322=0.015

d3=1.079-1.073=0.006 d4=1.229-1.233=-0.004

d5=0.936-0.934=0.002 d6=1.009-1.019=-0.01

d7=1.179-1.184=-0.005 d8=1.289-1.304=-0.015

Now we can calculate the mean of differences given by:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

And for the sample deviation we can use the following formula:

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

Describe the distribution of these differences using words. (which one is correct)

We can plot the distribution of the differences with the folowing code in R

differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)

hist(differences)

And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.

-The sample is too small to make judgments about skewness or symmetry.

Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)

$\bar X_{1}=1.173$ represent the mean for the operator 1

$\bar X_{2}=1.174$ represent the mean for the operator 2

$s_{1}=0.1506$ represent the sample standard deviation for the operator 1

$s_{2}=0.1495$ represent the sample standard deviation for the operator 2

$n_{1}=8$ sample size for the operator 1

$n_{2}=8$ sample size for the operator 2

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 means for the two groups are the same, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

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

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.

Calculate the statistic

We can replace in formula (1) like this:

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(14)}$

8 0

3 years ago