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Leya [2.2K]
2 years ago
9

Help please will give brainliest

Mathematics
1 answer:
dimulka [17.4K]2 years ago
4 0

Answer:

the black points on the top right marked with the letters are the answers

Hope it Helps

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The employees of two companies were asked how many times they checked their email during weekends. The table below shows the num
tekilochka [14]
Q1 of company A = 2.5
Q3 of company A = 8
Interquatile range = (Q3 - Q1)/2 = (8 - 2.5)/2 = 5.5/2 = 2.75

Q1 of company B = 2
Q3 of company B = 5.5
Interquatile range = (5.5 - 2)/2 = 3.5/2 = 1.75

Therefore, t<span>he interquartile range for Company A employees is 2 more than the interquartile range for Company B employees.</span>
3 0
2 years ago
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Dividing mixed numbers (pt. 2)
Reil [10]
A = 16 over 49 
b= 2 then 1 over 10
8 0
3 years ago
Please help me on number 4
Mariulka [41]
Angle 5 = 117° because it is vertical to angle 8
Angle 7 = 63° because it is symmetrical to angle 5
Angle 6 = 63° because it is vertical to angle 7.
Angle 1 = 117° because it is corresponding to angle 5.
Angle 2 = 63° because it is corresponding to angle 6.
Angle 3= 117° because it is corresponding to angle 8.
Angle 4 = 63° because it is corresponding to angle 7.
3 0
3 years ago
Mr. Park pays $1000 per month in rent. Before depositing his paycheck, his checking account balance was $359.47. After the depos
IrinaVladis [17]

Answer:

Mr Park has enough money to pay his rent.

Step-by-step explanation:

The amount of the monthly rent paid  = $1000

The account balance after depositing the paycheck = $1441.33

Now, to pay the rent,account should have more than the rent money $1000

Also, $1441.33  >  $1000

⇒ The amount in the bank after depositing the paycheck   >  The rent to be paid

Hence, Mr Park has enough money to pay his rent.

6 0
2 years ago
HELP!!!
xz_007 [3.2K]

Step-by-step explanation:

Given that the graph shows the normal distribution of the length of similar components produced by a company with a mean of 5 centimeters and a standard deviation of 0.02 centimeters.

A component is chosen at random, the probability that the length of this component is between 4.98 centimeters and 5.02

=P(|z|<1) (since 1 std dev on either side of the mean)

=2(0.3418)

=0.6826

=68.26%

The probability that the length of this component is between 5.02 centimeters and 5.04 centimeters is

=P(1<z<2) (since between 1 and 2 std dev from the mean)

=0.475-0.3418

=0.3332

=33.32%

6 0
3 years ago
Read 2 more answers
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