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3 years ago

You are scuba diving at

Mathematics

2 answers:

eduard3 years ago

7 0

-10 feet ok ok goody goody

devlian [24]3 years ago

3 0

Answer:

-10

Step-by-step explanation:

-8 feet minus 2 feet would equal -10 feet. It helps to think of it as a negative number instead, so it would be -8 ft + (-2) ft = -10 feet. Hope this helps!!

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Sigmund wrote four checks last month, and these were the only transactions for his checking account. According to his check regi

Lady bird [3.3K]

Answer:

The Check that hasn't been cleared yet is for $372.15.

Step-by-step explanation:

168.48+ 168.93+ 177.78+ 177.93= 693.12

887.79- 693.12= 194.67

1,065.27- 887.79= 177.48

194.67+ 177.48= 372.15

To check your answer:

372.15+ 693.12= 1,065.27

Hope this helped you!! (:

7 0

3 years ago

Read 2 more answers

What is the value of x a. 25 b. 12 c. 7 d. 5

tia_tia [17]

The answer is 5 because a^2 + b^2 = c^2

8 0

3 years ago

Read 2 more answers

I need help fast ....

Cloud [144]

Since x=-10, plug that into the equation.

f(x) = 2(-10) + 11
= -20 + 11
y = -9

So the ordered pair in (x,y) terms is (-10,-9).

4 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$