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3 years ago

After a storm, 135 trees out of 180 were left standing. What is the percentage loss of the number of trees?

Mathematics

1 answer:

beks73 [17]3 years ago

6 0

Answer:

Percentage loss of the number of trees is $25\%$ .

Step-by-step explanation:

Given: After a storm, $135$ trees out of $180$ were left standing.

To find: What is the percentage loss of the number of trees?

Solution:

We have,

Total number of trees $=180$

Number of trees left standing $=135$

Therefore, loss of trees $=180-135=45$

We now that $\text {Loss\%}=\frac{\text{Number of trees left}}{\text{Total number of trees}}\times 100 \%$

$\implies \text {Loss\%}=\frac{45}{180}\times 100 \%$

$\implies \text {Loss\%}=\frac{450}{18} \%$

$\implies \text {Loss\%}=25 \%$

Hence, the percentage loss of the number of trees is $25\%$ .

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Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

3 years ago

Help me plz for number 14 a-d

Svetllana [295]

The answer for number A would be 1/3

6 0

3 years ago

You own a motel and have a pricing structure that encourages rentals of rooms in groups.

zheka24 [161]

Answer:

$41

Step-by-step explanation:

One room rents for $68,
Two rooms for $65 each,
In general, the group rate per room is found by taking $3 off the base of $68 for each extra room.

So,

$a_1=\$68\\ \\a_2=\$65\\ \\a_3=\$62\\ \\...$

where $a_n$ is the rate per nth room rented.

This is an arithmetic sequence, so

$a_n=a_1+(n-1)\cdot d$

In your case,

$a_1=68\\ \\n=10\\ \\d=-3$

Hence

$a_{10}=68+(10-1)\cdot (-3)=68-9\cdot 3=68-27=41$

This means if a group rents 10 rooms, the charge per room is $41.

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3 years ago

What is the solution to 2/3a - 1/6 = 1/3

konstantin123 [22]

The solution would be: a = 3/4

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3 years ago

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A Web music store offers two versions of a popular song. The size of the standard version is 2.9 megabytes (MB). The size of the

Lady bird [3.3K]

Write the equations: $2.9s+4.7h=6076
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3 years ago