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Gala2k [10]
3 years ago
8

What is the 5th term in the pattern 1,-4,9,-16?

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
4 0
I think the answer is 25
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The sum of two numbers is equal to 70. Which of the following functions can be use to find the product of the two numbers if one
Temka [501]
Let x=number 1 let y=number2
there sum is equal to 70
then

number1+number2=70

substituting

x+y=70

y=70-x
let there product will be p
p=xy
substituting
p=x(70-x)
which is choice c

4 0
3 years ago
These are all my points I'm desperate please help I've posted this 4 other times
Juliette [100K]

Answer:

y = (x+3)^2 -1

Step-by-step explanation:

The vertex form of the equation is

y = a(x-h)^2 +k  where (h,k) is the vertex

The vertex is (-3,-1)

y = a(x- -3)^2 -1

y = a(x+3)^2 -1

Pick another point  (-2,0) and substitute it into the equation

0 = a(-2+3)^2 -1  to find a

0 =a(1)^2 -1

0 = a-1

1 = a

y = (x+3)^2 -1

8 0
3 years ago
Read 2 more answers
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
Help pls I'm rlly bad at math
Inessa05 [86]
The answer to the question is 4
3 0
3 years ago
Read 2 more answers
Convert 6/9 into a fraction with a denominator of 27.
alex41 [277]
Lets take 6/9. 9 is a multiple of 27, multiply 9 times 3 for a denominator of 27. Whatever we to do the bottom, we do to the top. So multiply 6 by 3 as well for numerator of 18. This gives us our answer, 18/27.
6 0
2 years ago
Read 2 more answers
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