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Art [367]
3 years ago
11

This probability distribution shows the typical grade distribution for a Geometry course with 35 students . A B С DF Grade Frequ

ency 5 10153 2 Find the probability that a student earns a grade of B. p=[?]

Mathematics
2 answers:
max2010maxim [7]3 years ago
8 0

Answer:

p=0.29

Step-by-step explanation:

hope it helps...

Nataly [62]3 years ago
4 0

Answer:

p=0.29

Step-by-step explanation:

Total ~ of ~students=35

Probability =10/35=0.29

<u>----------------------------</u>

hope it helps....

have a great day!!

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2. Find the midpoint of the segment with endpoints (1,4), (3,6)
I am Lyosha [343]

Answer:

2,5

Step-by-step explanation:

(1+3)/2, (4+6)/2

5 0
3 years ago
Solve the equation 4\pi =w-6\pi
maksim [4K]

Answer:

4/pi=w-6/pi

<u>4</u><u> </u><u> </u><u> </u>=<u> </u><u>w</u><u>-</u><u>6</u>

<u>Pi</u><u> </u><u> </u><u> </u><u> </u><u> </u><u>pi</u>

Cross mutiply

Pi(w-6)=4(pi)

Wpi - 6pi =4pi

Collect like terms

Wpi=4pi + 6pi

Wpi =10pi

<u>Divide</u><u> </u><u>both</u><u> </u><u>sides</u><u> </u><u>by</u><u> </u><u>pi</u>

<u>Wpi</u><u> </u>= <u>10pi</u>

Pi pi

W = 10

5 0
3 years ago
What is the solution to log 5 (x+30)= 3
Free_Kalibri [48]

Answer:

x = 95

Step-by-step explanation:

Using the rule of logarithms

• log_{b} x = n ⇔ x = b^{n}

Given

log_{5}(x + 30) = 3, then

x + 30 = 5³ = 125 ( subtract 30 from both sides )

x = 95

7 0
3 years ago
Julie bought a binder for $4 and several packs of paper that cost $0.70 each. If her total was $11.70, how many packs of paper d
timofeeve [1]
11 packs of paper is the answer
5 0
3 years ago
Suppose two $20 bills, three $10 bills, one $5 bill, and seven $1 bills are placed in a bag. If you were to pull a bill at rando
Korolek [52]

Answer:

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is \frac{7}{13}.

Step-by-step explanation:

From the given question, the bag contains;

$1 bill = 7

$5 bill = 1

$10 bill = 3

$20 bill = 2

Total number of bills in the bag = 13

Pulling a bill at random, the bills would have an expected value as follows:

For $1 bill, the expected value = \frac{7}{13}

For $5 bill, expected value = \frac{1}{13}

For $10 bill, expected value = \frac{3}{13}

For $20 bill, the expected value = \frac{2}{13}

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is \frac{7}{13}.

6 0
3 years ago
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