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2 years ago

\3 units

Mathematics

1 answer:

maxonik [38]2 years ago

3 0

Base of prism is seemed to be a triangle

Base=B=3
Height of base=H=6units
Height of prism=8

Area of base=

$\\ \sf\longmapsto \dfrac{1}{2}BH$

$\\ \sf\longmapsto \dfrac{1}{2}(3)(6)=3(3)=9units^2$

Now

Volume=V

$\\ \sf\longmapsto V=9(8)=72units^3$

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A seedless watermelon

MA_775_DIABLO [31]

$0.88

20%=0.2
0.2x4.4=0.88

4 0

3 years ago

What is 16,700,000,000,000,000 estimated as the product of a single digit and a power of 10

Vesna [10]

Hello!

Another way to put it would be scientific notation. to do so, you would need to get a digit 1-10 multiplied by 10 to a certain power.

For this problem, scientific notation is 1.67(10^16).

To check our answer, we will move the decimal to the right sixteen times because the exponent is positive. In other words, we will add fourteen zeros because we need two to turn 1.67 into 167. This gives us 16,700,000,000,000,000, which matches the question. (This number in word form is 16 quadrillion seven hundred trillion.)

Our final answer is 1.67(10^16).

I hope this helps!

6 0

4 years ago

suppose that a department contains 11 men and 17 women. how many different committees of 6 members are possible if the committee

slega [8]

If a department has 11 male employees and 17 female employees, then 198968 different committees of 6 members are possible with the condition that the committee has strictly more female employees than male.

As per the question statement, a department has 11 male employees and 17 female employees.

We re required to calculate the total number different committees that can be formed with 6 members strictly having more female employees than male.

Now for committee of 6 members to have more women than men, there can be two combinations:

(4 women and 2 men)

Or, (5 Women and 1 man).

That is, we will need to calculate the number of combinations we can have by selecting 4 female employees from a group of 17 and 2 Male employees from a group of 11 and the number of combinations we can have by selecting 5 female employees from a group of 17 and 1 Male employee from a group of 11 and add up the two number of combinations to obtain our required answer.

Then comes the most important thing to know to be able to solve this question, i.e., the formula to calculate combinations, which goes as

$nCr=\frac{n!}{r!(n-r)!}$

Therefore, the total number different committees that can be formed with 6 members strictly having more female employees than male is

$[(17C4)*(11C2)+(17C5)*(11C1)]\\=[(2380*55)+(6188*11)]\\=(130900+68068)\\=198968$

combination(s): In mathematics, a combination is a way of selecting items from a collection or set, where the order of selection does not matter, i.e., for example, we have a set of three numbers X, Y and Z and then, in how many ways can we select two numbers from each set, is defined by combination.

To learn more about Combinations, click on the link below.brainly.com/question/8044761

#SPJ4

7 0

2 years ago

Help ....,?.?.??.?.?..?.?.?.?

SVETLANKA909090 [29]

Answer:

Step-by-step explanation:

4 0

3 years ago

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

4 years ago