All categories

3 years ago

I will mark brainiest , if correct. please help.

Mathematics

2 answers:

suter [353]3 years ago

6 0

Area of the walls = 2*8*21 + 2*8*28 = 784 sq feet

so number of gallons required = 784/350 = 2.24 gallons answer

Vitek1552 [10]3 years ago

6 0

$S = 28 \cdot 21 \cdot 8 = 4704 \ ft^{2}. \newline n = \frac{4704}{350} = 14.$

You would need to buy 14 gallons of paint.

You might be interested in

What best describes the area of a polygon

never [62]

Answer:

The area of any regular polygon is given by the formula: Area = (a x p)/2, where a is the length of the apothem and p is the perimeter of the polygon. Plug the values of a and p in the formula and get the area. As an example, let's use a hexagon (6 sides) with a side (s) length of 10.

The area of a polygon is the two-dimensional set of all points surrounded by the sides of the polygon.

If you're looking for an equation, it varies based on the number of sides and the shape of the polygon.

Step-by-step explanation:

Apothem

A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). To find the area of a regular polygon, you use an apothem — a segment that joins the polygon’s center to the midpoint of any side and that is perpendicular to that side (segment HM in the following figure is an apothem).

6 0

3 years ago

4. Razi thinks of a number. He multiplies his number by 0.1, and then divides the

worty [1.4K]

The answer is 125!

Step-by-step explanation: 125 times 0.1 = 12.5 | 12.5 divided by 0.01 = 1250 | 1250 divided by 0.1 =1200!!

7 0

3 years ago

What is the product? (-2x-9y^2)(-4x-3)

telo118 [61]

Answer:-8x^2-6x+36xy^2+27y^2

Step-by-step explanation:

using the FOIL method

6 0

4 years ago

Read 2 more answers

Since 1900, the magnitude of earthquakes that measure 0.1 or higher on the Richter Scale in CA are distributed normally with a m

Gwar [14]

Answer:

a) 3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b) 1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c) 73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d) 99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e) 6.0735

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 6.2, \sigma = 0.5$

a.) What is the probability that a randomly selected earthquake in CA has a magnitude greater than 7.1?

This is 1 subtracted by the pvalue of Z when X = 7.1. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.1 - 6.2}{0.5}$

$Z = 1.8$

$Z = 1.8$ has a pvalue of 0.9641

1 - 0.9641 = 0.0359

3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b.) What is the probability that a randomly selected earthquake in CA has a magnitude less than 5.1?

This is the pvalue of Z when X = 5.1. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.1 - 6.2}{0.5}$

$Z = -2.2$

$Z = -2.2$ has a pvalue of 0.0139

1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c.) What is the probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1?

Now $n = 10, s = \frac{0.5}{\sqrt{10}} = 0.1581$

This is 1 subtracted by the pvalue of when X = 6.1. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{6.1 - 6.2}{0.1581}$

$Z = -0.63$

$Z = -0.63$ has a pvalue of 0.2643

1 - 0.2643 = 0.7357

73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d.) What is the probability that a ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

This is the pvalue of Z when X = 7.22 subtracted by the pvalue of Z when X = 5.7. So

X = 7.22

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.22 - 6.2}{0.1581}$

$Z = 6.45$

$Z = 6.45$ has a pvalue of 1

X = 5.7

$Z = \frac{X - \mu}{s}$

$Z = \frac{5.7 - 6.2}{0.1581}$

$Z = -3.16$

$Z = -3.16$ has a pvalue of 0.0008

1 - 0.0008 = 0.9992

99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e.) Determine the 40th percentile of the magnitude of earthquakes in CA.

This is X when Z has a pvalue of 0.4. So it is X when Z = -0.253.

$Z = \frac{X - \mu}{\sigma}$

$-0.253 = \frac{X - 6.2}{0.5}$

$X - 6.2 = -0.253*0.5$

$X = 6.0735$

6 0

3 years ago

Brad runs a lap every 5 minutes and Ty runs a lap every 4 minutes. If they started at the same time, when will they get to the b

olganol [36]

20 minutes. On brads 4th run and Ty’s 5th run they will be at the beginning at the same time.

5 0

3 years ago