Write the fractions from least to greatest 2/3 3/4 5/6

Four girls helped Mr.Day plant garden for their help hr gave the girls $24 to share equally later Mrs day gave each girl $2 for

katrin [286]

Answer:$8

Step-by-step explanation: there is 4 girls and $24 they split it equally so you would do 24/4 and that gives you 6. Then it says Mrs. Day gave each girl $2 so then you add 6+2 and then you get 8.

3 0

3 years ago

Read 2 more answers

How can the common difference between 375-315=30

Firlakuza [10]

Hello :
in the arithemitic sequence : an = an-1 + 30

3 0

3 years ago

What is a line graph?

scZoUnD [109]

A line graph, or line chart, shows info as a series of data points that are connected by line segments.

5 0

3 years ago

chubhunter [2.5K]

MNOP is similar to QRST with ratio of 5:4. That means the MP:QT ratio would also be 5:4

The length of QT would be
MP/QT = 5/4
85mm/QT=5/4
85mm*4= 5*QT
5QT= 340
QT=68mm

8 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