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3 years ago

7th grade simple solutions math 85-88

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

4 0

Answer:

-3 85-88 lets turn that around 88-85 is 3 but the problem is 85-88 so its -3

Step-by-step explanation:

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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,

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

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

I need help with this question

Viktor [21]

Y = 4x + 3 ---------(1)
y = -x - 2 ---------(2)

Subtract (2) from (1), you get
0 = 5x + 5
⇒ 5x + 5 = 0
⇒ 5x = -5
⇒ x = -5/5 = -1
y = -(-1) - 2 = -1

Answer is (-1,-1)

5 0

4 years ago

The following frequency table shows the number of cavities each of Dr. Freese's patients had.

Anit [1.1K]

Answer:

Step-by-step explanation:

13 patients had 0 cavities, according to the table! The righthand column is the frequency of people, and the left is the number of cavities they had.

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3 years ago

What is [b-5] over 8=5

julia-pushkina [17]

Answer:

B=45

Step-by-step explanation:

Multiply 8 by 5

Add 40 and 5 together

6 0

3 years ago

56% of what number is 84

damaskus [11]

We set up a proportion: 84/?= 56/100

Cross multiply: 56*?= 84*100
⇒ ?= 84*100/56= 150

56% of 150 is 84.

7 0

3 years ago