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4 years ago

What are the factors of 40 and 55? PLEASE SHOW WORK

Mathematics

1 answer:

Vladimir [108]4 years ago

8 0

40- 1-40 , 2-20, 4-10,5-8

55- 1-55, 5,11

factors are numbers times each other put together to get the number

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Help a uneducated sis out✊

N76 [4]

Answer & Step-by-step explanation:

The domain of a set of points refers to the input, also known as the x values. To find the domain, record all the x values given (x,y):

$domain={-2,0,3,5}$

:Done

**The range is the output, aka the y values.

5 0

3 years ago

Please I need help with this

Daniel [21]

180° in a triangle.
One angle is given 40°
Another angle can be found by 180-102=78°
180-78-40=62°
62+x=180
x=118°

3 0

2 years ago

Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning." These words begin a

IgorLugansk [536]

Answer: -8.16 to 15.84

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> is an interval in which we are a percentage sure the true mean is in the interval.

A confidence interval for a difference between two means and since sample 1 and sample 2 are under 30, will be

$x_{1}-x_{2}$ ± $t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$

where

x₁ and x₂ are sample means

t is t-score

$S_{p}$ is estimate of standard deviation

n₁ and n₂ are the sample numbers

The estimate of standard deviation is calculated as

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }$

where

s₁ and s₂ are sample standard deviation of each sample

Degrees of freedom is:

$df=n_{1}+n_{2}-2$

df = 12 + 9 - 2

df = 19

Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.

The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.

The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.

Calculating estimate of standard deviaton:

$S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }$

$S_{p}=\sqrt{\frac{299.07}{19} }$

$S_{p}=$ 15.74

The difference between means is

$x_{1}-x_{2}$ = 17.5 - 13.66 = 3.84

Calculating the interval:

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $27.21\sqrt{\frac{21}{108} }$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $27.21\sqrt{0.194}$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = 12

Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:

lower limit: 3.84 - 12 = -8.16

upper limit: 3.84 + 12 = 15.84

The interval is from -8.16 to 15.84.

5 0

3 years ago

Divide 575 by 14 by using partial quotients

levacccp [35]

575 / 14

14*10 = 140 -> 575 - 140 = 435 ......(1)
14*10 = 140 -> 535 - 140 = 295 ......(2)
14*10 = 140 -> 295 - 140 = 155 ......(3)
14*10 = 140 -> 155 - 140 = 15 ..... (4)

15 / 14 = 1 + Remainder 1

Number of 14 contained
10 + 10 + 10 + 10 + 1 = 41
Remainder 1

Answer 41 + Remainder 1

Verification: 41*10 + 1 = 574 + 1 = 575

6 0

4 years ago

If the length and width decrease by a factor of 3, what would be the effect on the area of the square?

Elenna [48]

Well, the affect would be 3x less on the area of the square

6 0

4 years ago

What are the factors of 40 and 55? *PLEASE SHOW WORK*

What are the factors of 40 and 55? PLEASE SHOW WORK