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3 years ago

14

Does anyone know the dilation of 4?

1 answer:

Mama L [17]3 years ago

8 0

Dilation of 4 = A

Hope this helps =3

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I need help and how do u get the answer to all of them

Vitek1552 [10]

It says to simplify so number 3 would be 12 $x^{3}$ $y^{4}$

Number 4 is $u^{24}$

Number 5 is $- \frac{7}{ c^{8} }$

4 0

2 years ago

Solve for brainliest

Evgesh-ka [11]

Answer:

7) 9.135x10^10

8) 3.428x10^-2

9) 2.5x10^-7

10) 4x10^10

Step-by-step explanation:

You’re only supposed to have one number to the left of the decimal point in scientific notation.

7) 9.135x10^10

8) 3.428x10^-2

9) 2.5x10^-7

10) 4x10^10

3 0

2 years ago

If y-18=14 what is the value of 3(y+5)?

svlad2 [7]

$y-18=14\ \ \ /+18\\\\y-18+18=14+18\\\\y=32\\\\substitute\ to\ 3(y+5):\\\\3\cdot(32+5)=3\cdot37=111$

4 0

2 years ago

GIVING BRAINLIEST PLEASE HELP

BabaBlast [244]

Answer:

S=16×2

S=32

................................

3 0

2 years ago

Read 2 more answers

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Portland. The stu

iVinArrow [24]

Answer:

$-0.061 < P_1 -P_2< 0.025$

Step-by-step explanation:

Give data:

$n_1 = 350$

$n_2 =250$

$x_1 = 23$

$x_2 = 21$

$\hat{P} 1 = \frac{x_1}{n_1} = \frac{23}{350} = 0.066$

$\hat{P} 2 = \frac{x_2}{n_2} = \frac{21}{250} = 0.084$

for 95% confidence interval

$\alpha = 1 - 0.95 = 0.05 and \alpha/2 = 0.025$

$z_{\alpha/2} = 1.96$ from standard z- table

confidence interval for P_1 and P_2 is

$\hat{P} 1 - \hat{P} 2 \pm z_{\alpja/2} \sqrt{\frac{\hat{P} 1(1-\hat{P} 1)}{n_1} +\frac{\hat{P} 2(1-\hat{P} 2)}{n_2}}$

$(0.066 - 0.084) \pm 1.96 \sqrt{\frac{0.066(1-0.066)}{350} +\frac{0.084(1-0.084)}{250}}$

$-0.018 \pm 0.043$

confidence interval is

$-0.018 - 0.043 < P_1 -P_2$

$-0.061 < P_1 -P_2< 0.025$

7 0

2 years ago