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EastWind [94]
3 years ago
10

7. Look at this figure

Mathematics
2 answers:
aliya0001 [1]3 years ago
6 0
B is your answer!!!

Hope it helps
kap26 [50]3 years ago
3 0

Answer:

thx for the 32 points

Step-by-step explanation:

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1. Team 7C is going on a field trip. The team has 25 students in each homeroom. Each homeroom is allotted 2 vans and 1 car. 5 st
Anna [14]

Answer:

10 Kids In Each Van.

Why?

You already know that 5 students will be in 1 car. That leaves 20 students that need to be assigned to vans. Divide 20 by 2 and you get 10!

4 0
3 years ago
I’ll mark Brainiest?
Usimov [2.4K]
The answer is D. Group =2
8 0
3 years ago
A store is having a sale on jelly beans and almonds. For 2 pounds of jelly beans and 5 pounds of almonds, the total cost is $10
pickupchik [31]
 For this question it is going to be 80 $.
7 0
3 years ago
The work of a student to solve the equation 4(2x − 4) = 8 + 2x + 8 is shown below:
Setler [38]

Step-by-step explanation:

Here, The work of a student to solve the equation :

4(2x-4)=8+2x+8

The steps for calculating the value of x are :

Step 1: 4(2x − 4) = 8 + 2x + 8  

Step 2: 6x − 8 = 16 + 2x  

Step 3: 6x − 2x = 16 + 8  

Step 4: 4x = 24  

Step 5: x = 6

Step 2 is incorrect, as the student instead of multiplying, adds the number as :

4(2x − 4) = 8 + 2x + 8  

Here, the correct step is :

8x − 16 = 16 + 2x

6 x = 32

x = 5.33

So, the error is in STEP 2 and the correct option is (b).

4 0
3 years ago
Read 2 more answers
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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