Which property can be used to justify the conclusion?

What is the circumference of this circle with a diameter of 15cm? Use π ≈ 3.14. Round your answer to the nearest tenth of a cent

Yakvenalex [24]

The answer is 49.00 after rounding it to the nearest tenths

6 0

4 years ago

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A rectangle is 3 times as long as it is wide, and its perimeter is 80 centimeters. Find its dimensions

Free_Kalibri [48]

Say the width is W and the length is L

L=3W (and of course W=W)

Now you use the perimeter formula P=2L+2W

80=2(3W) + 2W)

80=6W+2W

80=8W Divide both sides by 8

W=10

Now that you know the width, substitute it back into L=3W to find the length

L=3(10)

L=30

The dimensions are length: 30 cm and width: 10 cm

8 0

4 years ago

3x - 5 = 10 x = ______

Sphinxa [80]

Answer:

3x - 5 = 10 x

The answer is: $x=-\frac{5}{7}$

Step-by-step explanation:

3x - 5 = 10x

First, Please, Add 5 to both sides, Which would be:

3x - 5 + 5 = 10x + 5

Second, Please, Simplify your answer, Which would be:

3x = 10x + 5

Third, Please, Subtract 10x from both sides, Which would be:

3x - 5 = 10x = 10x + 5 - 10x

Forth, Please, Simplify your answer, Which would be:

-7x = 5

Fifth, Please, Divide both sides by -7, Which would be:

$\frac{-7x}{-7}=\frac{5}{-7}$

Last and Finally, Simplify your answer, Which would be:

$x=-\frac{5}{7}$

I hope I helped :)

5 0

4 years ago

Residences of U.S. Citizens According to the U.S. Census, 67.5% of the U.S. population were born in their state of residence. In

trasher [3.6K]

Answer:

P(X < 126) = 0.0418

Step-by-step explanation:

To solve this question, we use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Sample of 203, 67.5% of the U.S. population were born in their state of residence.

This means that $n = 203, p = 0.675$

So, for the approximation:

$\mu = E(X) = np = 203*0.675 = 137.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{203*0.675*0.325} = 6.67$

What is the probability that fewer than 126 were born in their state of residence?

Using continuity correction, this is P(X < 126 - 0.5) = P(X < 125.5), which is the pvalue of Z when X = 125.5. So

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

$Z = \frac{125.5 - 137.025}{6.67}$

$Z = -1.73$

$Z = -1.73$ has a pvalue of 0.0418

So

P(X < 126) = 0.0418

8 0

3 years ago

What does "PEMDAS" stand for?

Fed [463]

P= parentheses or ()
E= exponents such as $x^{2}$
M= multiplying
D= division
A= Addition
S= Subtraction

Another popular acronym is BEDMAS, which is pretty much the same except for division and multiplication are switched and it uses the word Brackets to describe these: ().

8 0

3 years ago

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