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DaniilM [7]
3 years ago
10

Math help me please you will be mark as the brainiest I do not have much time left

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
7 0
A little tip: you should put the question in next time :)!
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PLEASE DO PLEASE
MA_775_DIABLO [31]

Answer: 8x^4.5, 4x^3


Step-by-step explanation:


5 0
3 years ago
2x+4=8 what would the equation be and what does x equal
Dafna11 [192]

Answer:

x=2

Step-by-step explanation:

2x+4=8

2x=8-4

2x=4

x=4/2

x=2

4 0
3 years ago
Read 2 more answers
How many \frac{1}{2} 2 1 ​ - pound packages of cheese can the deli make with 1212 pounds of cheese?
monitta

Given:

Total cheese = 12 pound

Cheese in each package = \dfrac{1}{2} pounds

To find:

The number of cheese packages.

Solution:

We need to divide the total amount of cheese by quantity of cheese in each package to find the the number of cheese packages.

\text{Number of cheese packages}=\dfrac{\text{Total cheese}}{\text{Cheese in each package}}

\text{Number of cheese packages}=\dfrac{12}{\dfrac{1}{2}}

\text{Number of cheese packages}=12\times \dfrac{2}{1}

\text{Number of cheese packages}=24

Therefore, there are 24 number of cheese packages.

5 0
3 years ago
Find the value of x. Round your answer to the nearest tenth. Show your work please!
netineya [11]
To find x we can use that fact that x is in the adjacent side of the 22° angle. We also know that 11 is our hypotenuse. Now, we can use trigonometry to find x.

cos(Ф)=adjacent/hypotenuse
cos(22°)=x/11
11 · cos(22°)= x/11 · 11    multiplicative inverse
11 cos (22°)=x
10.2≈x
3 0
3 years ago
2. At the test center giving the exam from the previous problem, the firm needs to plan for proper staffing. Historical data sug
kozerog [31]

Answer:

a) I would use the following distribution fir the amount of applicants arriving in a 20 minute period

P_X(k) = \frac{e^{-1.5} \, * \, 1.5^k}{k!}

b) In a 50 hour period, we will expect to need 0.4679*50 = 23.3948 hours with the extra staff member

Step-by-step explanation:

a) For a total amount of arrivals in a 20 minute period i would use a Poisson distribution with parameter λ = 1.5 (the average). The distribution X is given by this formula

P_X(k) = \frac{e^{-1.5} \, * \, 1.5^k}{k!}

b) For one hour, the average will be 1.5*60/20 = 4.5 applicants. The distribution Y for the amount of applicants in one hour is given by the following formula

P_Y(k) = \frac{e^{-4.5} \, * \, 4.5^k}{k!}

First, we want to find the probability of Y being greater then or equal to 5. We can obtain the probability of the complementary event and substract it from 1. That event is equal to the probability of Y being equal to 0,1,2,3 or 4

P(Y=0) = e^{-4.5}

P(Y=1) = e^{-4.5}* 4.5

P(Y=2) = e^{-4.5} * 4.5²/2 = e^{-4.5} * 10.125

P(Y=3) = e^{-4.5} * 4.5³/6 = e^{-4.5} * 15.1875

P(Y=4) = e^{-4.5} * 4.5⁴/24 = e{-4.5} * 17.08594

Thus,

P(Y < 5) = P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4) = e^{-4.5} * (1+4.5+10.125+15.1875+17.08594) = 0.5321

Therefore,

P(Y ≥ 5) = 1-0.5321 = 0.4679

In a 50 hour period, we will expect to need 0.4679*50 = 23.3948 hours with the extra staff member.

5 0
4 years ago
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