If a T-shirt cost $12.50 and you have a coupon for 30% off how much will the T-shirts be if you use the coupon

Solve for x in the equation x2+2x+ 1 = 17. x=-1± √15 x=-17/17 x=-2+2.15 X=-1+13

larisa86 [58]

Answer:

x=12, but if x2 is an exponent, x=1+ the square root of 17.

Step-by-step explanation:

4 0

3 years ago

How much longer is 2 1/2 inch long piece than a 2/5 inch long piece of string

tino4ka555 [31]

two and one tenth longer

3 0

3 years ago

Solve this inequality for the y variable: -5x + 3y> 15

SCORPION-xisa [38]

Answer:

y>5/3x+5

Step-by-step explanation:

You want to treat this inequality as if it were a regular equation to solve for y or x.

First you want to get y by itself by getting rid of -5x. Add 5x on both sides of the inequality to then get 3y>5x+15. Lastly, divide on both sides of the inequality by three to isolate y and you then get the inequality y>5/3x+5

5 0

3 years ago

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 229 wi

meriva

Answer:

A. 99% CI for p: 0.125 < p < 0.259

B. 99% CI for p: 0.807 < p < 0.911

C. 95% CI for p: 0.776 < p < 0.844

D. 95% CI using t-distribution: 23.7 < µ < 40.3

Step-by-step explanation:

A.

$p=\frac{x}{n} =\frac{44}{229} = 0.192139738$

Confidence = 99% = 0.99

$\alpha$ =1 - confidence = 1 - 0.99 = 0.01

Critical z-value = $z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.192139738(1-0.192139738)}{229} }$

$SE=\sqrt{\frac{0.155222059}{229} }$

SE = 0.026035

E = Margin of Error

E = z-value * SE = 2.58 * 0.026035

E = 0.067170516

Lower Limit = p - E = 0.192139738 - 0.067170516 = 0.125

Upper Limit = p + E = 0.192139738 + 0.067170516 = 0.259

99% CI for p: 0.125 < p < 0.259

B.

$p=\frac{x}{n} =\frac{256}{298} = 0.8590604$

Confidence = 99% = 0.99

$\alpha$ =1 - confidence = 1 - 0.99 = 0.01

Critical z-value = $z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.8590604(1-0.8590604)}{298} }$

$SE=\sqrt{\frac{0.121075629}{298} }$

SE = 0.020157

E = Margin of Error

E = z-value * SE = 2.58 * 0.020157

E = 0.052004

Lower Limit = p - E = 0.8590604 - 0.052004 = 0.807

Upper Limit = p + E = 0.8590604 + 0.052004 = 0.911

99% CI for p: 0.125 < p < 0.259

C.

$p=\frac{x}{n} =\frac{405}{500} = 0.81$

Confidence = 95% = 0.95

$\alpha$ =1 - confidence = 1 - 0.95 = 0.05

Critical z-value = $z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.81(1-0.81)}{500} }$

$SE=\sqrt{\frac{0.1539}{500} }$

SE = 0.017544

E = Margin of Error

E = z-value * SE = 1.96 * 0.017544

E = 0.034387

Lower Limit = p - E = 0.81 - 0.034387 = 0.776

Upper Limit = p + E = 0.81 + 0.034387 = 0.844

95% CI for p: 0.776 < p < 0.844

D.

x = sample mean = 32

s = sample standard deviation = 13

n = sample size = 12

Confidence = 95% = 0.95

$\alpha$ =1 - confidence = 1 - 0.95 = 0.05

Being the sample size less than 30, we use the statistical t.

df = degree of freedom = n - 1 = 12 - 1 = 11

Critical t-value = $t_{\alpha/2,df}=t_{0.05/2,11}=t_{0.025,11}=2.201$ (from t-table, two tails, df=11)

SE = standard error

$SE=\frac{s_{x} }{\sqrt{n} } =\frac{13 }{\sqrt{12} }=3.752777$

E = margin of error

E = t-value * SE = 2.201 * 3.752777 = 8.259862

Lower Limit = x - E = 32 - 8.259862 = 23.7

Upper Limit = x + E = 32 + 8.259862 = 40.3

95% CI using t-distribution: 23.7 < µ < 40.3

Hope this helps!

4 0

3 years ago

3. What is the volume of the shape? Round to the nearest hundredth. (TT=3.14)

s344n2d4d5 [400]

Answer:

2814.87

Step-by-step explanation:

Formula = π * r (squared) * height

Radius = 8cm

Height = 14cm

Answer = π*64*14, or 2814.87 (to the nearest hundredth)

7 0

3 years ago