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34kurt
3 years ago
11

Solve for b2 in A = h ( b1 + b2 ), if A = 16, h = 4, and b1 = 3.

Mathematics
2 answers:
nlexa [21]3 years ago
8 0
Divide 4 by both sides
Then you get 4=3+b2
Then subtracts 3 from four
You get 1=b2

pychu [463]3 years ago
5 0

Answer:

b2 = 5

Step-by-step explanation:

16 = 1/2 x 4 (3 + b2)

16 = 2 (3 + b2)

16 = 6 + 2b2

16-6 = 2b2

10 = 2b2

10/2 = 2b2/2 (the 2s canceled out)

5 = b2

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3 years ago
Marissa has twice as much money as Frank. Christina has $20 more than Marissa. If Christina has $100, how much money does Frank
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Answer:

a. Take a grid that shows represents the amount of money Frank has,

Say F,

( shown below ),

b. Since, Marissa has twice as much money as Frank,

So, the amount Marissa has = 2F

If M represents the amount Marissa has,

M = 2F

c. Now, Christina has $20 more than Marissa,

Add 20 to the grid M

d.  The below diagram provides enough information to determine the value of the variable m

e. The amount Christina has = M + 20 = 2F + 20

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2F + 20 = 100

2F = 80

F = 40

Hence, the amount frank has = $ 40,

Amount Marissa has = 2(40) = $ 80,

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3 years ago
A recycling truck begins its weekly route at the recycling plant at point A, as pictured on the coordinate plane below. It trave
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78

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7 0
3 years ago
You order seventeen burritos to go from a Mexican restaurant, seven with hot peppers and ten without. However, the restaurant fo
Kisachek [45]

Answer:

P=0.369

Step-by-step explanation:

In this binomial probability case, the success is to pick a burrito with hot peppers, there are 7 out of 17, the probability is p=7/17=0.4118. The formula is

P=_nC_kp^k(1-p)^{n-k}

Where n is the number of burritos taken at random, k is the number of success (burrito with hot peppers), p is the probability of success. _nC_k denotes the number of combinations and its formula is:

_nC_k=\frac{n!}{k!(n-k)!}

At least two burritos have hot peppers means having two or three burritos, these two cases are calculated and sum to get the probability of at least two burritos that have hot peppers.

P=_nC_kp^k(1-p)^{n-k}=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}

P_2=\frac{3!}{2!(3-2)!}(0.4118)^2(1-0.4118)^{3-2}=\frac{3\times2!}{2!1!} (0.1696)(0.5882)^1\\P_2=3 (0.1696)(0.5882)=0.2993

P_2=\frac{3!}{3!(3-3)!}(0.4118)^3(1-0.4118)^{3-3}=\frac{1}{0!} (0.0698)(0.5882)^0\\P_2=1 (0.0698)(1)=0.0698

P=P_2+P_3=0.2993+0.0698=0.3691

Round to three decimal places: P=0.369

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