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Lubov Fominskaja [6]
3 years ago
15

A person invested $810 in an account growing at a rate allowing the money to double

Mathematics
2 answers:
Hitman42 [59]3 years ago
5 0

Answer:

10.6

Step-by-step explanation:

Andrew [12]3 years ago
3 0

Answer:

i dont know bro

Step-by-step explanation:

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Assume that y varies inversely
Marina86 [1]

Answer:

y = 2

Step-by-step explanation:

y varies inversely with x setup is:

y = k/x

7 = \frac{k}{2/3}  (find 'k')

k = 7/1 · 2/3

k = 14/3

use what you know about 'k' and 'x' to solve for 'y'

y = 14/3 ÷ 7/3  (remember to multiply by the reciprocal when dividing fractions)

y = 14/3 · 3/7

y = 2

3 0
3 years ago
At a local river there were 48 alligators laying on the riverbank. if 5/6of the alligators were sleeping how many were awake
qaws [65]
8 alligator because is u divide 48/5 you get 40 and the remaining that r awake is 8
7 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 is the imaginary part of the number 9?​
KengaRu [80]

Answer:

<h2>0i</h2>

Step-by-step explanation:

The imaginary number has form:

<em>a + bi</em>

<em>a</em><em> - real part</em>

<em>bi</em><em> - imaginary part</em>

<em>i</em><em> - imaginary unit (i = √-1)</em>

We have the number 9.

<em>9</em><em> is a real part</em>

An imaginary part is equal 0.

Therefore the imaginary part of number 9 is 0i.

4 0
3 years ago
What precent of 44 is 11 <br><br>helllllllllllllllllllllllllllllllllllllllllllllllllllllllllp<br>plz
olganol [36]

Answer:

25.

Step-by-step explanation:

44:11*100 = 25

5 0
2 years ago
Read 2 more answers
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