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3 years ago

What numbr must be subtracted from -20 to obtain 15?

Mathematics

1 answer:

Radda [10]3 years ago

8 0

Answer:

-5

Step-by-step explanation:

-5--20

=-5+20

=20-5

=15

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3y-4=2 solve using the properties of equality

pav-90 [236]

Answer:

3y_4=2

3y=2+4

3y=6

y=6/2

y=2

3 0

3 years ago

Here is a pattern made from sticks:

oee [108]

Answer:

maybe like 12

Step-by-step explanation:

I would think 12 because because it is 2 of the square so I think 6 + 6 correct me of I'm wrong.

5 0

3 years ago

One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe

sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% 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 people who disagree with the statement = $\frac{1812}{2625}$ = 0.69

n = sample of respondents = 2625

p = population proportion of people who disagree with statement

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ , $0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ ]