All categories

2 years ago

The ages (in years) of the eight passengers on a bus are listed below.

1 answer:

5 0

11 22 29 35 40 46 91

22 24.5 25.5 29

24.5 + 25.5 = 50

50/2 = 25

honestly i was confused because the amount of numbers is odd = 7 so i thought that the median would obviously be 35 since it’s inn the middle but it’s not one of the listen numbers

so i thinks it’s C

You might be interested in

I had $5.00. My mom gives me $10.00, My dad gave me $30.00. My aunt and uncle give $100.00. I had another $5.00. How much money

Sladkaya [172]

You HAD $10 but now you have $150

3 0

3 years ago

Can someone please help me ASAP

forsale [732]

Answer:

its B

Step-by-step explanation:

8 0

3 years ago

If y varies jointly as x and the cube of z and y=16 when x=4 and z=2 then y=0.5 when x=-8 and z=-3?

Evgesh-ka [11]

Answer: F (False)

Step-by-step explanation:

Jointly variation has the following form:

y=kxz

Where k is a constant of propotionality.

Substitute values:

If y=16, x=4 and z=2, then k is:

$16=k(4)(2)^{3}\\k=1/2$

If x=-8 and z=-3 the the value of y is:

$y=(1/2)(-8)(-3)^{3}\\y=-108$

Then the answer is FALSE.

8 0

3 years ago

Read 2 more answers

Penelope is going to the carnival to ride the rides. It costs $20 to get into the

Dimas [21]

Answer:

35=0.5x+20

Step-by-step explanation:

y=mx+b

35= money you can spend

0.5= cost of each ride

x=number of rides you ride

20= price to enter

5 0

3 years ago

2. The National Safety Council routinely analyzes the benefit of seat belt use on driver safety. Their data showed that among 28

JulijaS [17]

Answer:

We conclude that there is difference in the proportion of deaths between the 2 groups.

Step-by-step explanation:

We are given that among 2823 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed.

Let $p_1$ = proportion of deaths when drivers were not wearing seat belts.

$p_2$ = proportion of deaths when drivers were wearing seat belts.

So, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no difference in the proportion of deaths between the 2 groups}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is difference in the proportion of deaths between the 2 groups}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of deaths when drivers were not wearing seat belts = $\frac{31}{2823}$ = 0.011

$\hat p_2$ = sample proportion of deaths when drivers were wearing seat belts = $\frac{16}{7765}$ = 0.002

$n_1$ = sample of drivers not wearing seat belts = 2823

$n_2$ = sample of drivers wearing seat belts = 7765

So, the test statistics = $\frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }$

= 4.438

The value of z test statistics is 4.438.

Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is difference in the proportion of deaths between the 2 groups.

Also, Margin of error (E) = $1.96 \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }$

= $1.96 \times \sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} }$

= 0.00397

5 0

3 years ago