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2 years ago

11

WILL CHOOSE BRAINLIEST

1 answer:

algol [13]2 years ago

4 0

Answer:

This is the answer if any problem just ask me

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If As income is 20% more than Bs income, how much percent is Bs income less than As income

vitfil [10]

$As\ is\ 120\%\ Bs\\\\120\%=1.2\to\frac{As}{Bs}=1.2\\\\\frac{As}{Bs}=\frac{12}{10}\to\frac{Bs}{As}=\frac{10}{12}\to\frac{Bs}{As}=\frac{5}{6}\\\\\frac{5}{6}\cdot100\%=\frac{500}{6}\%=83\frac{1}{3}\%\\\\100\%-83\frac{1}{3}\%=16\frac{2}{3}\%\\\\Answer:Bs\ income\ is\ 16\frac{2}{3}\%\ less\ than\ As\ income.$

6 0

3 years ago

If two pyramids have the same height, what must be true of the pyramids for them to also have the same volume? The pyramids must

vodka [1.7K]

Answer:

The area of the bases must be the same.

Step-by-step explanation:

Volume=LxWxH

Length and width create the base so another formula would be V=B•h, and if they have the same height they would need the same base. Hope this helps, sorry if I’m incorrect.

6 0

2 years ago

Read 2 more answers

2/3 of ___=6<br> Can you help me?

Artyom0805 [142]

9 is the answer because 1/3 = 3

6 0

2 years ago

Read 2 more answers

Select the correct answer.

Nina [5.8K]

Answer:

D. $x^2 + 3x - 88$

Step-by-step explanation:

Area of rectangle = $length (l) * width (w)$

Length of rectangle = (x - 8)

Width = (x + 11)

Area of rectangle = $(x - 8)(x + 11)$

Expand the expression

$x(x + 11) -8(x + 11)$ (distributive property of multiplication)

$x^2 + 11x - 8x - 88$

Combine like terms

$x^2 + 3x - 88$

Expression for the area of the rectangle = $x^2 + 3x - 88$

7 0

2 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

2 years ago