Question 6

The total weight of the golf balls from Julio’s golf bag was 273 grams. If each golf ball weighed 45.5 grams, which graph shows

MariettaO [177]

ANSWER:

A

EXPLAIN:

Since all the six balls have same weight and the total weight is given as 273g, then weight of one ball will be

273/6=45.5 g

1 ball=45.5 g

2 balls =2*45.5=91 g

3 balls =3*45.5=136.5 g

4 balls=4*45.5=182 g

5 balls =5*45.5=227.5 g

6 balls =6*45.5=273 g

7 0

3 years ago

PLZZZZZZZZ HELP MEEEEE

lukranit [14]

To the first question it would be a because all of the brown spots look like an equal in size equation be with number two because according to the ox

7 0

3 years ago

PLEASE ANSWERRRRRRRRRRRRRRR

hoa [83]

Let $T,S$ be the number of days spent at Tahoe and San Francisco, respectively.

We don't know the values of $T$ and $S$ yet, but we know that the holiday lasted 9 days:

$T+S=9$

We also know that each day spent in Tahoe costed 350 and each day spent in San Francisco costed 475. So, the total cost of the holiday is the sum of the number of days muliplied by their cost:

$350T+475S=3650$

If we put the two equations together, we have the system

$\begin{cases}T+S=9\\350T+475S=3650\end{cases}$

Which yields

$S=4,\quad T=5$

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

Hi, in need of a little assistance.Question: calculate the internal area of the living room.

Dominik [7]

The living room is rectangular. We would find the area by applying the formula for calculating the area of a rectangle which is expressed as

Area = length x width

From the information given,

internal length = 5.2

internal width = 2.95

Thus,

internal area of the living room = 5.2 x 2.95

internal area of the living room = 15.34 m^2

7 0

1 year ago