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4 years ago

7

a school purchsed basketball for $20 each and footballs for $15 each. they a brought a total of 28 for $480 how many of each typ

e did they buy. define yur varaibles, write a system. solve the system. PLZZZ ANSWER I NEED IT FOR A TEST AND ITS TIME LIMIT

1 answer:

Contact [7]4 years ago

3 0

Answer:

13

Step-by-step explanation:

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A triangle, ABC, has side lengths of 15 in, 28 in and 17 in

ikadub [295]

Answer:

17/28

if you're solving for cosine

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Suppose the doormat has a design on it that divide it into squares measuring 1/10 meter by 1/10 meter how many squares is the do

amm1812

There are 10 squares and each square is a meter

3 0

3 years ago

) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa

asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

<u><em>Let </em></u> $p_1$ <u><em> = population proportion of drivers 16-24 years old who buckle up .</em></u>

<u><em /></u> $p_2$ <u><em> = population proportion of drivers 25-69 years old who buckle up .</em></u>

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no significant difference in seat belt use}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a difference in seat belt use}

The test statistics that would be used here <u>Two-sample z proportion statistics;</u>

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 drivers 16-24 years old who buckle up = 79%

$\hat p_2$ = sample proportion of drivers 25-69 years old who buckle up = $\frac{924}{1100}$ = 84%

$n_1$ = sample of 16-24 years old drivers = 1000

$n_2$ = sample of 25-69 years old drivers = 1100

So, <u><em>test statistics</em></u> = $\frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }$

= -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u><em>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics 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 </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that there is a difference in seat belt use.

4 0

3 years ago

Solve this step by step

Kaylis [27]

Step-by-step explanation:

the formula to calculate a circumference is

$x \times \pi$

(consider x as the diameter)

so I'll be noting the circumference as C alongside the circle's ranking

a.

$ca = 6 \times 22 \div 7 \\ = 44 \div 7$

b.

$cb = 7 \times 22 \div 7 \\ = 154 \div 7 \\ = 22$

c.

$cc \: = 4 \times 22 \div 7 \\ = 88 \div 7$

hope I helped

8 0

3 years ago

A sum amount <br>to rupees 34476 in two and a half years at 4% CI. the sum is

Kruka [31]

Given:

Amount = Rs. 34476

Rate of compound interest = 4%

Time = $2\dfrac{1}{2}=2.5$ years

To find:

The principal value.

Solution:

Formula for amount is

$A=P\left(1+\dfrac{r}{100}\right)^t$

Where, P is principal value, r is rate of interest and t is time in years.

Putting the given values, we get

$34476=P\left(1+\dfrac{4}{100}\right)^{2.5}$

$34476=P\left(1+0.04\right)^{2.5}$

$34476=P\left(1.04\right)^{2.5}$

$\dfrac{34476}{\left(1.04\right)^{2.5}}=P$

Now,

$P=31256.0090$

$P\approx 31256$

Therefore, the value of sum or principal value is Rs.31256.

6 0

3 years ago