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Gala2k [10]
3 years ago
12

The perimeter of the square at the right is 48 inches. What is the area of the square at the right? Explain how you found your a

nswer.
Mathematics
1 answer:
Masja [62]3 years ago
4 0
If you have an image attached, i cant see it. Maybe just repost the question again cause i reallu cant see the attached image
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20 points free<br> How to make a cow bark
yKpoI14uk [10]

Answer:

RUFF RUFF RUFF RUFF RUFF

6 0
3 years ago
Solve for x ....... thanks !
Len [333]
If we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b

16=2⁴
32=2⁵
rememeber that (x^m)^n=x^{mn}

16^{3x+2}=32^{-2x-7}
(2^4)^{3x+2}=(2^5)^{-2x-7}
2^{4(3x+2)}=2^{5(-2x-7)}
2=2 so we conclude that 4(3x+2)=5(-2x-7)

4(3x+2)=5(-2x-7)
expand/distribute
12x+8=-10x-35
add 10x both sides
22x+8=-35
minus 8 both sides
22x=-43
divide both sides by 22
x=-43/22
8 0
3 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

7 0
3 years ago
Bobcat park is a rectangular park with an area of 5 1/5 square miles.Its width is 1 19/20 miles. How long is the park
mel-nik [20]

area = length times width, so length = area / width

5 1/5 mi^2 5.2 mi^2

Here, length = ------------------- = --------------- = 2.66666... mi = 2 2/3 mi (answer)

1 19/20 mi 1.95 mi

4 0
3 years ago
-(1 - 9x) - 8 helpppp meeeeeee
madreJ [45]

Answer

9x - 9

Step-by-step explanation

sorry im late

7 0
3 years ago
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