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2 years ago

What are the answers

Mathematics

2 answers:

miss Akunina [59]2 years ago

7 0

Answer:

see below

Step-by-step explanation:

1) 18

2) 45

3) 75

4) 4

5) 7

6) 14

7) 3

8) 9

9) 135

10) 8

11) 60

12) 15

13) 72

14) 45

BartSMP [9]2 years ago

6 0

The top answers are answered correctly

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Erick paid a $75 registration fee to join the gym. He pays $15 each month.

liraira [26]

Answer:

$150

Step-by-step explanation:

Erick has to pay $15 per month so for 5 months he will have to pay $75 because 15 times 5 is 75. So $75 for registration fee + $75 for 5 months to use the gym give us $150

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2 years ago

If z varies directly as x and z=30 when x=8 find z when x=4

dangina [55]

Z=kx, 30=8k, so k=30/8=15/4 and z=15x/4, so when x=4, z=15.

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3 years ago

Which shows the expression below in simplified form?

statuscvo [17]

Answer:

B. 9x10^4

Step-by-step explanation:

$\frac{4.5 \times {10}^{8} }{5 \times {10}^{3} } \\ = 0.9 \times {10}^{8 - 3} \\ = 0.9 \times {10}^{5} \\ = 9 \times {10}^{4}$

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3 years ago

A bag of marbpes contains 6 blue marbles, 2 yellow marbles, 4 red marbles, and 1 green marbpe. What is the probability of reachi

klio [65]

Answer:

1/6 or 1.6666666

Step-by-step explanation:

12 marbles, 2 yellow

12/2= 6

1/6

5 0

2 years ago

Read 2 more answers

An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results an

igomit [66]

Answer:

Mean = 1.57

Variance=0.31

Step-by-step explanation:

To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:

1) Both surgeries are successful (X=2).

$P(e_1)=0.90*0.67=0.603$

2) Left knee unsuccessful and right knee successful (X=1).

$P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067$

3) Right knee unsuccessful and left knee successful (X=1).

$P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297$

4) Both surgeries are unsuccessful (X=0).

$P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033$

Then, the mean can be calculated as the expected value:

$M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57$

The variance can be calculated as:

$V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111$

3 0

3 years ago