Marie invests $25,000 in a 60-day term deposit that pays four percent simple interest per year.

Goryan [66]

Well, when you want to make 15 a percentage, You need to know of what, for an example, 20 percent of 100 is 20% percent because 20/100= 0.20 which equals 20%. What is 15 of? Fill in the blank. 15 percent of ___=? which would be 15/?

4 0

3 years ago

If veronica reed is paid a gross weekly salary of $1036.00, what is her gross annual salary?

Ber [7]

So, (weekly) she receives 1036.00.
The question is asking to find out the (annual) salary which is a year.
There are about 52 weeks in a year.
To find the amount being payed annually we multiply 1036 by 52
=53,872.

3 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 = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

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