What is the value of x in the equation 25x2 = 16?

35 Percent of 70 is 24.5
We assume, that the number 70 is 100% - because it's the output value of the task. We assume, that x is the value we are looking for. If 70 is 100%, so we can write it down as 70=100%. We know, that x is 35% of the output value, so we can write it down as x=35%. Now we have two simple equations: 1.)70=100% 2.)x=35%where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that: 70/x=100%/35% Now we just have to solve the simple equation, and we will get the solution we are looking for.

3 0

3 years ago

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A website sells a dress at £40 and shoes at £32. The website has an offer: Buy a dress and shoes together and get 1/2 off the pr

tekilochka [14]

Answer:

$\pounds36$

Step-by-step explanation:

Cost of one dress = $\pounds40$

Cost of one pair of shoes = $\pounds32$

Cost of one dress and a pair of shoes is $40+32=\pounds72$

They get half off the price because they buy both

$\dfrac{72}{2}=\pounds32$

The shipping and handling charge is $\pounds4$ .

So, the total amount Ruth has to pay is $32+4=\pounds36$ .

6 0

3 years ago

The image of the point (1, -2) after a rotation of 180° about the origin is

Harlamova29_29 [7]

Answer:

(-1,2)

Step-by-step explanation:

Found answer on brainly.

4 0

3 years ago

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The amount of money spent on textbooks per year for students is approximately normal.

Contact [7]

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.101 < $t_1_8$ < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the 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 students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions

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

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]