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3 years ago

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15% of Mr.Rileys classes watch the local news. 136 of his students do not watch the local news. How many students are in Mr.Rile

ys classes

1 answer:

Helga [31]3 years ago

8 0

(100-15)% = 85% = 136 students

136 / (85%) = 130*(100/85) = 160 students

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Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago

12x^2+ 16x factor by gcf

qaws [65]

Answer:

The factor of 12x^2 + 16 is 4x (3x+4)

7 0

2 years ago

56 out of 154 in simplest form of a fraction

marissa [1.9K]

We start with 56/154. Dividing each number by two, we have:

28/77

Dividing each number by seven, we have:

4/11

As the greatest common factor of 4 and 11 is 1, this fraction is in simplest form. Thus, 56/154 is equal to 4/11.

4 0

3 years ago

Read 2 more answers

The area of a circular oil slick on the surface of the sea is increasing at a rate of [150m^2/s] how fast is the radius changing

Akimi4 [234]

Answer with Step-by-step explanation:

We are given that

$\frac{dA}{dt}=150m^2/s$

a.Radius =25 m

We have to find rate of change of radius.

We know that

Area of circle, $A=\pi r^2$

Differentiate w.r.t time

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Substitute the values then we get

$150=2\pi (25)\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{150}{50\pi}=\frac{3}{\pi} m/s$

$\frac{dr}{dt}=\frac{3}{\pi} m/s$

b.A= $1000m^2$

Substitute the values then we get

$1000=\pi r^2$

$\pi=3.14$

$r^2=\frac{1000}{3.14}$

$r=\sqrt{\frac{1000}{3.14}}=17.8m$

$\frac{dA}{dr}=2\pi r\frac{dr}{dt}$

Substitute the values then we get

$150=2\pi(17.8)\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{150}{2\pi(17.8)}=\frac{150}{35.6\pi}$ m/s

$\frac{dr}{dt}=\frac{75}{17.8\pi}$ m/s

4 0

3 years ago

Simplify the expression /-9/(3-2i)+(1+5i)

Elodia [21]

36+27i because /-9/ equals 9 so you distribute 9 in the parentheses then you add like terms

7 0

3 years ago