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katovenus [111]
3 years ago
10

For a 100 ponits : 13+(-19)= ANSWER:

Mathematics
2 answers:
ahrayia [7]3 years ago
5 0

Answer:

the answer for me is 13+(-19)= -6

Step-by-step explanation:

-6

KATRIN_1 [288]3 years ago
3 0

Answer: -6

Step-by-step explanation:

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Pls answer this with an explanation if you can
Nuetrik [128]

Answer:

The answer is D

Step-by-step explanation:

$13.95 divided by 100 = 1% which is equal to $0.1395

To get 15% of her bill, you need to multiply $0.1395 by 15 which gives you $2.0925. You then round this to 2 significant figures, which totals to $2.10.

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3 years ago
4. A random variable X has a mean of 10 and a standard deviation of 3. If 2 is added to each value of X, what will the new mean
Ede4ka [16]

Adding 2 to each value of the random variable X makes a new random variable X+2. Its mean would be

E[X+2]=E[X]+E[2]=E[X]+2

since expectation is linear, and the expected value of a constant is that constant. E[X] is the mean of X, so the new mean would be

E[X+2]=10+2=12

The variance of a random variable X is

V[X]=E[X^2]-E[X]^2

so the variance of X+2 would be

V[X+2]=E[(X+2)^2]-E[X+2]^2

We already know E[X+2]=12, so simplifying above, we get

V[X+2]=E[X^2+4X+4]-12^2

V[X+2]=E[X^2]+4E[X]+4-12^2

V[X+2]=(V[X]+E[X]^2)+4E[X]-140

Standard deviation is the square root of variance, so V[X]=3^2=9.

\implies V[X+2]=(9+10^2)+4(10)-140=9

so the standard deviation remains unchanged at 3.

NB: More generally, the variance of aX+b for a,b\in\mathbb R is

V[aX+b]=a^2V[X]+b^2V[1]

but the variance of a constant is 0. In this case, a=1, so we're left with V[X+2]=V[X], as expected.

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3 years ago
A is between C and D. True or false?
DanielleElmas [232]

Answer:

false

Step-by-step explanation:

false , because A is between C and B , it must be on the same straight line

3 0
3 years ago
Last week’s and this week’s low temperatures are shown in the table below. Low Temperatures for 5 Days This Week and Last Week L
STatiana [176]

Question:

Temperatures for 5 Days This Week and Last Week

Low Temperatures This Week (Degrees Fahrenheit) 4 10 6 9 6 Low Temperatures Last Week (Degrees Fahrenheit) 13 9 5 8 5 Which measures of center or variability are greater than 5 degrees? Select three choices.

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

d)the mean absolute deviation of this week’s temperatures

e) the mean absolute deviation of last week’s temperatures

Answer:

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

Step-by-step explanation:

I would be verifying the options a, b, and c in my answer above through calculations which are shown below.

We were given the following data:

Low Temperatures This Week (Degrees Fahrenheit)

4, 10, 6, 9, 6

Low Temperatures Last Week (Degrees Fahrenheit)

13, 9, 5, 8, 5

We are to find which measures of center or variability are greater than 5 degrees.

Option a

The mean of this week’s temperatures

(4+ 10+ 6+9+ 6) °F ÷ 5 = 35 °F ÷5 = 7°F

Option a is correct because it measures of center or variability which is 7 °F is higher than 5°F

Option b

The mean of last week’s temperatures

(13+ 9 + 5 + 8 + 5) °F = 40°F ÷ 5 = 8°F

Option b is correct , because its measure of variability which is 8°F is greater than 5°F.

Option c

the range of this week’s temperatures

This week's temperature is given as

(4, 10, 6, 9, 6) °F

Range is defined as the difference between the highest number and the lowest number

Range of this week's temperature = (10 - 4) °F = 6°F

Hence Option c is correct because it measures of center or variability which is 6°F is greater than 5°F

From the above calculations we can accurately confirm that options a, b, and c are correct because the measures of their center or variability is greater than 5°F

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3 years ago
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