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Monica [59]
3 years ago
11

How did they get 72?

Mathematics
2 answers:
Anon25 [30]3 years ago
8 0

Answer: 12x6=72 get it?

Step-by-step explanation:

lyudmila [28]3 years ago
6 0

Answer:

they multiplied all the sides

Step-by-step explanation:

You might be interested in
What is the slope of the line perpendicular to the segment with endpoints (2,0) and (-6,3)?
r-ruslan [8.4K]

Answer:

A line that passes (2, 0) and (-6, 3) has slope:

S1 = (-6 -2)/(3 - 0) = -4/3

Another line that is perpendicular with above line has slope:

S2 = -1/S1 = -1/(-4/3) = 3/4

=> Option B is correct

Hope this helps

:)

3 0
3 years ago
The Lacrosse booster club is holding a raffle for a fundraiser. They will sell 100 tickets for $5 each and select 4 winners. All
Nadusha1986 [10]

Answer:

<h2>There are 3,921,225 ways to select the winners.</h2>

Step-by-step explanation:

This problem is about combinations with no repetitions, because the same person can't win four times. It's a combinaction because the order of winning doesn't really matter.

Combinations without repetitions are defined as

C_{n}^{r}  =\frac{n!}{r!(n-r)!}

Where n=100 and r=4.

Replacing values, we have

C_{100}^{4}  =\frac{100!}{4!(100-4)!}=\frac{100!}{4! 96!}=\frac{100 \times 99 \times 98 \times 97 \times 96!}{4! \times 96!}=  \frac{94,109,400}{24}= 3,921,225

Therefore, there are 3,921,225 ways to select the winners.

Additionally, as you can imagine, the probability of winning is extremely low, it would be 3,921,225 to 1.

3 0
3 years ago
A recent study done by the National Retail Federation found that 2019 back-to-school spending for all US households who have sch
MissTica

Answer:

Step-by-step explanation:

Hello!

The working variable is:

X: Back-to-school expense of a US household with school-aged children.

X~N(μ;σ²)

μ= $697

σ= $120

a. What is the probability that 2019 back-to-school spending for a US household with school-aged children is greater than $893?

Symbolically: P(X>$893)

First, you standardize the probability using Z= (X-μ)/σ ~N(0;1)

P(X>$893)= P(Z>(893-697)/120)= P(Z>1.63)

To resolve this question you have to use the table of cumulative probabilities for the standard normal distribution. These tables accumulate probabilities from the left, symbolically P(Z≤Z₀), so to reach probabilities greater than a Z₀ value you have to subtract the cumulative probability until that value from the maximum probability value 1:

P(Z>1.63)= 1 - P(Z≤1.63)= 1 - 0.94845= 0.05155

b. Provide the Z-score corresponding to the 2019 back-to-school spending of $1,200, and the probability of 2019 back-to-school spending for a household with school-aged children is less than $1,200.

P(X<$1200) = P(Z<(1200-697)/120)= P(Z<4.19)= 1

According to the empirical rule of the normal distribution, 99% of the data is between μ ± 3σ. This, logically, applies to the standard normal distribution. Considering that the distribution's mean is zero and the standard deviation is one, then 99% of the probabilities under the standard normal distribution are within the Z values: -3 and 3, values below -3 will have a probability equal to zero and values above 3 will have probability equal to one.

c. Find Q3 (Third Quartile).

Q3 in the value that marks three-quarters of the distribution, in other words, it has 75% of the distribution below it and 25% above, symbolically:

P(Z≤c)=0.75

In this case, you have to look in the center of the right Z-table (positive) for the probability of 0.75 and then the margins to find the Z-score that belongs to that cumulative probability:

c= 0.674

Now you reverse the standardization to see what value of X belongs to the Q3:

c= (X-μ)/σ

X= (c*σ)+μ

X= (0.674*120)+697= $777.88

d. Find Q1 (First Quartile)

To resolve this you have to follow the same steps as in c., just that this time you'll look for the value that marks the first quarter of the distribution, symbolically:

P(Z≤d)= 0.25

In this case, since the probability is below 0.5 you have to look for the Z value in the left table (negative).

d= -0.674

d= (X-μ)/σ

X= (d*σ)+μ

X= (-0.674*120)+697= $616.12

e. What is the value of the IQR for the distribution of 2019 back-to-school spending for a US household with school-aged children?

IQR= Q3-Q1= $777.88 - $616.12= $161.76

f. Interpret the value of the IQR from question 2e within the context of the problem.

$161.76 represents the distance between 75% of the Back-to-school expense of a US household 25% of the Back-to-school expense of US households.

g. What is the proportion of 2019 back-to-school spending within 1.50 standard deviations of the mean?

"Within 1.50 standard deviations of the mean" can be symbolized as "μ ± 1.5σ" or "μ - 1.5σ≤ Z ≤μ + 1.5σ"

P(μ - 1.5σ≤ Z ≤μ + 1.5σ)

Since the mean is zero and the standard deviation is one:

P(-1.5 ≤ Z ≤ 1.5)= P(Z≤1.5) - P(Z≤-1.5)= 0.933 - 0.067= 0.866

h. What is the 2019 back-to-school spending amount such that only 3% of households with school-age children spend more than this amount?

The "top" 3% means that you are looking for a value of the variable that has above it 0.03 of probability and below it 0.97%, first you look for this value under the standard normal distribution and then you reverse the standardization to reach the corresponding value of the variable:

P(Z>h)= 0.03 ⇒ P(Z≤h)=0.97

h= 1.881

h= (X-μ)/σ

X= (h*σ)+μ

X= ( 1.881*120)+697= $922.72

i. Which US household is more unusual, a US household with back-to-school spending of $600 or a US household with back-to-school spending of $900?

Under this kind of distribution, the "most usual" values are around the center (near the mean) and the "unusual" values will find themselves in the tails of the Gaussian bell.

To check which one is more unusual you have to see their distance with respect to the mean.

(X-μ)/σ

(600-697)/120= -0.8083

(900-697)/120= 1.69

An expense of $900 is more unusual than an expense of $600 (600 is almost the expected expenses)

j. Let's say the Smith family spent $815 on buying school supplies this fall. Provide an interpretation of the Smith family's 2019 back-to-school spending, i.e. what can you say about the percentage of all other US households with school-age children that have higher back-to-school spending than the Smith family?

P(X>$815) = P(Z>(815-697)/120)= P(Z>0.98)

1-P(Z≤0.983)= 0.837

83.7% of the families will have back-to-school expenses of $815 or more.

I hope it helps!

6 0
3 years ago
URGENT!! Really need help on this test
Katyanochek1 [597]

Answer:

G= L &M

H=L &M

K=P&N

J=P&N

7 0
3 years ago
Help me please this test is very important !
Lilit [14]

Answer:

d

Step-by-step explanation:

5>2n+11

5-11>2n

-6>2n

-6/2>n

n<or equal to -3

bubble is filled pointing left

6 0
3 years ago
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