An athlete runs 8 miles in 50 minutes on a treadmill. At this rate, how long will it take the athlete to run 9 miles?

I need help plz !!!!!!!

ArbitrLikvidat [17]

The answer is 12.3 cents. Hope this helped :))

7 0

3 years ago

Find the total amount due on a loan of $5,000 for 8 years at 10% simple interest.

nata0808 [166]

10% of 5000 is 500 and 500 times 8 is 4000 so the total amount due would be 9000 :)

3 0

3 years ago

If you save up $100 every 2 weeks, how much money would you have in a year ?

Pachacha [2.7K]

alr! so we know that there are around 52 weeks in a year (not counting leap years). and you save up 100 bucks every 2 weeks. then to find the answer, we do...

100(52/2)

100(26)

= $2600

youre welcome!

8 0

3 years ago

A homing pigeon flew a distance of 120 miles in 4 days. If it flew x miles in the first 3 days, write an expression to tell how

mart [117]

If the last day is y. and then the answer will be y=120-x

8 0

3 years ago

Read 2 more answers

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed

tatyana61 [14]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 8 minutes

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes

(b) The P-value of the test statistics is 0.0436.

(c) We conclude that the actual mean waiting time equals the standard at 0.05 significance level.

(d) 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Step-by-step explanation:

We are given that the length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes.

A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation of 3.2 minutes.

Let $\mu$ = actual mean waiting time.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 8 minutes {means that the actual mean waiting time does not differs from the standard}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes {means that the actual mean waiting time differs from the standard}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

So, test statistics = $\frac{8.5-8}{\frac{3.2}{\sqrt{120} } }$

= 1.71

The value of t test statistics is 1.71.

(b) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.9564 = 0.0436

(c) Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual mean waiting time equals the standard.

(d) Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population proportion, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $8.5-1.96 \times{\frac{3.2}{\sqrt{120} } }$ , $8.5+1.96 \times{\frac{3.2}{\sqrt{120} } }$ ]

= [7.93 minutes , 9.07 minutes]

Therefore, 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Yes, it support our above conclusion.

3 0

4 years ago