Please please help me i’ll give you brainliest

In a group of Year 9 students, the ratio of boys to girls is 9 : 7. There are 4 more boys than girls in this group. How many stu

KATRIN_1 [288]

Answer: 32

Step-by-step explanation:

There are 18 boys and 14 girls (4 more boys than girls. 32 students total.

7 0

3 years ago

How many ways can Gary, Cayla, and

Kryger [21]

Answer:

6

Step-by-step explanation:

G,C,P

C,P,G

P,G,C

G,P,C

P,C,G

C,G,P

4 0

4 years ago

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir

Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, we find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

$\sigma = \sqrt{484} = 22$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$5 = 1.96*\frac{22}{\sqrt{n}}$

$5\sqrt{n} = 43.12$

$\sqrt{n} = \frac{43.12}{5}$

$\sqrt{n} = 8.624$

$(\sqrt{n})^{2} = (8.624)^{2}$

$n = 74.4$

We need a sample size of at least 75.

6 0

3 years ago

Read 2 more answers

The equation shows the cost, c, for a group attending a basketball game based on the number of people, p.How many people could a

Black_prince [1.1K]

Step-by-step explanation:

Read the note harder

5 0

3 years ago

Rectangle ABCD has vertices at (-7,-2); (1, -2); (1, -8); and (-7, -8) respectively. If GHJK is a similar rectangle where G(2, 5

Gennadij [26K]

Answer:

Points J and K could be located at:

J(2,2)

K(6,2)

Step-by-step explanation:

Consider the vertices have a x and y coordinate:

A: x coordinate=-7 y coordinate=-2

B: x coordinate=1 y coordinate=-2

C: x coordinate=1 y coordinate=-8

D: x coordinate=-7 y coordinate=-8

G: x coordinate=2 y coordinate=5

H: x coordinate=6 y coordinate=5

Then it is possible to calculate the distance between the x and y coordinates:

x coordinate of Vertices AB:

x coordinate of B- x coordinate of A=1-(-7)=8

The distance between A and B is 8

y coordinate of Vertices AC:

y coordinate of A- y coordinate of C=-2-(-8)=6

The distance between A and C is 6

Then we know that the side AB of Rectangle ABCD measures 8 and the side AC, measures 6.

Repeat the analysis with Rectangle GHJK. In this case, is only possible to calculate the distance with x coordinate.

x coordinate of Vertices GH:

x coordinate of H- x coordinate of G=6-(2)=4

The distance between G and H is 4

We can see that the distance in x of the Rectangle ABCD is 8, and the distance in x of the Rectangle GHJK is 4, it means that the measure of ABCD is twice GHJK.

Then, if the distance in y coordinate of Vertices AC is 6, we could say that the distance in y coordinate of Vertices GJ is 3.

Points J and K could be located at:

J(2,2)

K(6,2)

3 0

3 years ago