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inna [77]
3 years ago
11

I need help fast!!!!!

Mathematics
2 answers:
Vesna [10]3 years ago
7 0
-48 is the answer to the question

mean = Summation of Data/Number of Data Added

mean = [(-54)+(-32)+(-70)+(-25)+(-65)+(-42)] / 6

mean = -288 / 6
mean = -48
sammy [17]3 years ago
7 0

Answer:  The required mean is -48.

Step-by-step explanation:  We are given to find the mean of the following integers :

-54, -32, -70, -25, -65, -42.

We know that

the mean of a given set of integers is equal to the sum of the integers divided by the number of integers.

Therefore, the mean of the given set of integers is

M\\\\\\=\dfrac{-54+(-32)+(-70)+(-25)+(-65)+(-42)}{6}\\\\\\=\dfrac{-54-32-70-25-65-42}{6}\\\\\\=\dfrac{-288}{6}\\\\=-48.

Thus, the required mean is -48.

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3 years ago
What is the factored form of x^3 + 125?​
klemol [59]

Answer:

(X+5)(x^2-5x+25)

Step-by-step explanation:

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A water sprinkler has a range of 5 meters as shown. The jet of water from the sprinkler sweeps out at an angle of 85° as the noz
telo118 [61]
I found the image that accompanied this problem.
We need to solve for the area of the sector.

A = n/360 * π * r²
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6 0
3 years ago
Read 2 more answers
What are the solutions of the quadratic equation (x + 3)(x +3) = 49? Ax = -2 and x = -16 Bx = 2 and x = -10 Cx = 4 and x = -10 D
Ivahew [28]

Answer:

4 and -10

Step-by-step explanation:

\displaystyle (x + 3)(x +3) = 49 \\ x^2+3x+3x+9=49 \\ x^2 +6x+9=49 \\ x^2 + 6x + 9 - 49 = 0 \\x^2+6x-40=0 \\\\ \Delta=b^2-4ac \\ \Delta=6^2-4 \cdot 1 \cdot (-40) \\ \Delta=36+160 \\ \Delta=196 \\ \\ X_{1,2}=\frac{-b \pm \sqrt{\Delta} }{2a}  \\ \\ X_1=\frac{-b+\sqrt{\Delta} }{2a} = \frac{-6+14}{2} = \frac{8}{2}=4 \\ \\ X_2=\frac{-b-\sqrt{\Delta} }{2a}  = \frac{-6-14}{2}=\frac{-20}{2} = -10

4 0
3 years ago
A sample size of n = 64 is drawn from a population whose standard deviation is o = 5.6.
AlekseyPX

Answer:

The margin of error for a 99% confidence interval for the population mean is 1.8025.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

In this problem:

\sigma = 5.6

So

M = 2.575*\frac{5.6}{\sqrt{64}} = 1.8025

The margin of error for a 99% confidence interval for the population mean is 1.8025.

5 0
3 years ago
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