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3 years ago

Find the area and circumference of 4

Mathematics

1 answer:

wolverine [178]3 years ago

8 0

The answer would be A=50.27m^2

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SOLVE THE SYSTEM OF EQUATIONS −2x+6y=−38 3x−4y=32

WARRIOR [948]

X= 4 and Y =-5 im not really 100% sure but its all i know

6 0

3 years ago

Read 2 more answers

Factor the expression 49b²-36

Travka [436]

Answer:

49b² - 36

Step-by-step explanation:

Find the factors of 49 & 36:

49 = 1, 7, 49

36 = 1, 2, 6, 18, 36

As you can tell, there are no common factors (1 does not count), therefore, the expression given is already in most factored form.

49b² - 36 is your answer.

7 0

3 years ago

Simplify the following fractions 8/24 8/32 6/30 12/24 9/21 5/20

Mademuasel [1]

Answer:

for 8/24 the answer will be 1/3.

and for 8/32 the answer will be 1/4.

and for 6/30 the answer will be 1/5.

and for 12/24 the answer will be 1/2.

and for 9/21 the answer will be 3/7.

and for 5/20 the answer will be 1/4

Step-by-step explanation:

4 0

3 years ago

A group of dental researchers are testing the effects of acidic drinks on dental crowns. They have five containers of crowns lab

mestny [16]

Answer:

$\begin{array}{cccccc}{x} & {V} & {W} & {X} & {Y} & {Z} & P(x) & {0.20} & {0.20} & {0.20} & {0.20} & {0.20} \ \end{array}$

Step-by-step explanation:

Given

$S = \{V,W,X,Y,Z\}$

$n(S) = 5$

Required

The probability model

To do this, we simply calculate the probability of each container.

So, we have:

$P(V) = \frac{n(V)}{n(S)} = \frac{1}{5} = 0.20$

$P(W) = \frac{n(W)}{n(S)} = \frac{1}{5} = 0.20$

$P(X) = \frac{n(X)}{n(S)} = \frac{1}{5} = 0.20$

$P(Y) = \frac{n(Y)}{n(S)} = \frac{1}{5} = 0.20$

$P(Z) = \frac{n(Z)}{n(S)} = \frac{1}{5} = 0.20$

So, the probability model is:

$\begin{array}{cccccc}{x} & {V} & {W} & {X} & {Y} & {Z} & P(x) & {0.20} & {0.20} & {0.20} & {0.20} & {0.20} \ \end{array}$

5 0

2 years ago

Which of the following statements is not true? a) The standard deviation of the sampling distribution of sample mean = σ/√n b) T

I am Lyosha [343]

Answer:

e) The mean of the sampling distribution of sample mean is always the same as that of X, the distribution from which the sample is taken.

Step-by-step explanation:

The central limit theorem states that

"Given a population with a finite mean μ and a finite non-zero variance σ2, the sampling distribution of the mean approaches a normal distribution with a mean of μ and a variance of σ2/N as N, the sample size, increases."

This means that as the sample size increases, the sample mean of the sampling distribution of means approaches the population mean. This does not state that the sample mean will always be the same as the population mean.

6 0

3 years ago