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BabaBlast [244]
3 years ago
12

Linda and Imani are each traveling in a car to the beach. Linda’s travel is modeled by a table. Imanis travel it modeled by grap

h.

Mathematics
1 answer:
tino4ka555 [31]3 years ago
7 0

Answer:

Imani traveled faster because her distance surpassed Linda's distance after the fourth hour.

Step-by-step explanation:

1. Understand the slope:

      Ok, so the slope basically tells us how fast Imani and Linda traveled, so if we solve for the slope we know how fast Imani and Linda traveled.

2. Find Linda's slope:

      - Take any two points: (0,10) and (1, 65).

      Slope = (<u>y1 - y2)/(x1 - x2)</u>

<u> </u>                 = (65-10)/(1-0)

                   = 55

    Since the slope is 55, we know her speed is 55 miles per hour.

2. Find Imani's slope:

      - Take any two points: (0,0) and (1, 60).

      Slope = (<u>y1 - y2)/(x1 - x2)</u>

<u> </u>                 = (60-0)/(1-0)

                   = 60

    Since the slope is 60, we know her speed is 60 miles per hour.  

3. Compare: Since 60>55, we know Imani traveled faster. If we look at options, we can see there's only one answer which shows that Imani traveled faster, so that must be the answer.

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Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test
BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\  \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}

- Applying the formula, we have:

\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\  \\ t=\frac{5}{3.1752} \\  \\ t\approx1.5747 \end{gathered}

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

7 0
1 year ago
Solve: (6x2 + 5x + 1) + (x + 2).
Mademuasel [1]

Answer:

6x+6x2+3

Step-by-step explanation:

6x2+5x+1+x+2

Combine 5x and x to get 6x.

6x2+6x+1+2

Add 1 and 2 to get 3.

6x2+6x+3

7 0
2 years ago
The weight of people in a small town in Missouri is known to be normally distributed with a mean of 186 pounds and a standard de
OleMash [197]

Answer:

the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

Step-by-step explanation:

The summary of the given statistical data set are:

Sample Mean = 186

Standard deviation = 29

Maximum capacity 3,417 pounds or 17 persons.

sample size = 17

population mean =3417

The objective is to determine the probability  that a random sample of 17 persons will exceed the weight limit of 3,417 pounds

In order to do that;

Let assume X to be the random variable that follows the normal distribution;

where;

Mean \mu = 186 × 17 = 3162

Standard deviation = 29* \sqrt{17}

Standard deviation = 119.57

P(X>3417) = P(\dfrac{X - \mu}{\sigma}>\dfrac{X - \mu}{\sigma})

P(X>3417) = P(\dfrac{3417 - \mu}{\sigma}>\dfrac{3417 - 3162}{119.57})

P(X>3417) = P(Z>\dfrac{255}{119.57})

P(X>3417) = P(Z>2.133)

P(X>3417) =1- 0.9834

P(X>3417) =0.0166

Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

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3 years ago
What place is eight and 46.38
svetoff [14.1K]
Hundreds place!
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3 years ago
Can someone please help?
PIT_PIT [208]

Step-by-step explanation:

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4 0
3 years ago
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