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zlopas [31]
3 years ago
10

Evaluate. 10squared +6⋅7+8 70 142 150 190

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
7 0

Answer:

the answer is 150. this is thw answer

Viefleur [7K]3 years ago
5 0

666.7

Google helps but if you think through it and read in between the lines it makes since and do it when pencil and paper to think it out.

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A rectangle has a height of 3p2+1 and a width of p3+4 your answer should be a polynomial in standard form
drek231 [11]

Answer:

3p^5+p^3+12p^2+4

Step-by-step explanation:

5 0
3 years ago
Y = 2 - 6x<br>17 y - x=1 solve the system​
Bond [772]

Answer:

(33/103,8/103)

In point form it is (33/103,8/103)

In equation form it is x = 33/103, y = 8/103

Hope this helps!

Consider giving brainliest. I only have 4 brainliest left and then my rank will go up.

7 0
3 years ago
When and image is dilated, the area of the new image is the [blank] of the [blank] of the original figure and the [blank] factor
finlep [7]

Answer: ummmmmmmmmmm smatrofhdkbdjbd

Step-by-step explanation:

5 0
3 years ago
A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after
lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                         P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of persons who were still employed primarily as engineers  = \frac{111}{200} = 0.555

           n = sample of graduates = 200

           p = population proportion of engineers

<em>Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                 significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } , 0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } ]

 = [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

7 0
3 years ago
What's 9/5 divided by 14/15? im stuck
Varvara68 [4.7K]

Answer:

When you divide fractions, you multiply by the reciprocal, So, you would have 9/5 * 15/14 you can cancel,  and get 9/1 * 3/14, and end up with 27/14,

5 0
2 years ago
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