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3 years ago

How do u do 10 to the 3 power

Mathematics

2 answers:

sammy [17]3 years ago

8 0

It's (10x10x10) which is 1000

earnstyle [38]3 years ago

5 0

10 times 10 times 10. It equals 1000

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ABC preschool spun $1,475 on diapers this year.if the diapers cost $25 per box how many boxes did they used?

Aleks04 [339]

Cost = number of objects x price per object

$1,475 = x boxes * $25

x = 1,475/25 = 59 boxes of diapers

7 0

3 years ago

The length of a rectangle is twice the width. the perimeter is 32 cm more than the width. find the dimensions of the rectangle.

SCORPION-xisa [38]

Width: x
Length: 2x
Perimeter=2x+2x+x+x=6x
32 cm more than width means x+32
6x=x+32
5x=32
x=6.4
Dimension is 6.4×12.8 (width×length)

7 0

3 years ago

James earns $2.20 less than twice gregs hourly wage. If james earns $16.50 per

Cerrena [4.2K]

Greg’s hourly wage is $9.35 per hour

5 0

3 years ago

Read 2 more answers

PLEASE ANSWERRRRRRRRRRRRR

Sunny_sXe [5.5K]

Answer:

y = 23x-82

Step-by-step explanation:

The equation is written in the form y = mx+b

We need to find the slope

m = (y2-y1)/(x2-x1)

= (79-56)/(7-6)

= 23/1

The slope is 23

y= 23x +b

Put in a point for x and y to determine b

10 =23(4) +b

10 =92 +b

Subtract 92 from each side

10-92 = 92-92 +b

-82 =b

The equation is

y = 23x-82

3 0

3 years ago

The scores of individual students on the American College Testing (ACT) Program College Entrance Exam have a normal distribution

tekilochka [14]

Answer:

The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

Step-by-step explanation:

Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);

$P(0\leq X\leq 19.0)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{19-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}, \mu=18.6, \sigma=6.0$

$P(-3.1\leq Z\leq 0.0667)=\Phi(0.0667)-\Phi (-3.1)=\Phi(0.0667)-(1-\Phi (3.1))=0.52790+0.99903-1=0.52693$

P(Xi≥19.0)=0.473

$\{Yi=0, Xi< 19\\Yi=1, Xi\geq 19\}$

p=0.473

Yi~Ber(0.473)

$P(\frac{1}{n}\displaystyle\sum_{i=1}^{n}X_i\geq 19)=P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{n}X_i\~{}N(n\mu, \sqrt{n}\sigma),\displaystyle\sum_{i=1}^{400}X_i\~{}N(7440, 372)$

Then:

$P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)=1-P(0$

$P(\displaystyle\sum_{i=1}^{n}Y_i=1)=P(\displaystyle\sum_{i=1}^{400}Y_i=1)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{400}Y_i\~{}N(400\times 0.473, \sqrt{400}\times 0.499)=\displaystyle\sum_{i=1}^{400}Y_i\~{}N(189.2; 9.98)$

$P(\displaystyle\sum_{i=1}^{400}Y_i=1)\~{=}P(0.5$

Then:

the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

7 0

3 years ago