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jekas [21]
3 years ago
14

If one worker can assemble 9 products per hour, and another worker can assemble 6 products per hour, how long will it take them

to assemble 50 products if they both start working at the same time
Mathematics
1 answer:
leva [86]3 years ago
3 0

Answer:

  3 hours 20 minutes

Step-by-step explanation:

Together, the workers can assemble 9 + 6 = 15 products per hour. So the assembly of 50 products will take ...

  (50 products)/(15 products/hour) = 50/15 hours = 3 1/3 hours

The two workers can assemble 50 products in 3 1/3 hours.

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Which of the graph of f(x)=x-1/x^2-x-6
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Answer:

a on edge

Step-by-step explanation:

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Which would be the best name for a function that takes the size of an engine and returns the maximum speed of a car?
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Answer:

Size

Step-by-step explanation:

Just got the answer out

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Four teams A,B,C, and D compete in a tournament, and exactly one of them will win the tournament. Teams A and B have the same ch
beks73 [17]

Answer:

A= 0,2

B= 0,2

C= 0,4

D=0,2

Step-by-step explanation:

We know that only one team can win, so the sum of each probability of wining  is one

P(A)+P(B)+P(C)+P(D)=1

then we Know that the probability of Team A and B are the same, so

P(A)=P(B)

And that the  the probability that either team A or team C wins the tournament is 0.6, so P(A)+Pc)= 0,6, then P(C)= 0.6-P(A)

Also, we know that team C is twice as likely to win the tournament as team D, so P(C)= 2 P(D) so P(D) = P(C)/2= (0.6-P(A))/2

Now if we use the first formula:

P(A)+P(B)+P(C)+P(D)=1

P(A)+P(A)+0.6-P(A)+(0.6-P(A))/2=1

0,5 P(A)+0.9=1

0,5 P(A)= 0,1

P(A)= 0,2

P(B)= 0,2

P(C)=0,4

P(D)=0,2

4 0
3 years ago
In the parking lot shown, the lines
jolli1 [7]
22 is the answer because I just did it
8 0
4 years ago
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5
Elodia [21]

Answer:

7.3% percentage of the bearings produced will not be acceptable.

Step-by-step explanation:

Consider the provided information.

Average diameter of the bearings it produces is .500 inches. A bearing is acceptable if its diameter is within .004 inches of this target value.

Let X is the normal random variable which represents the diameter of bearing.

Thus, 0.500-0.004<X<0.500+0.004

0.496<X<0.504

The bearings have normally distributed diameters with mean value .499 inches and standard deviation .002 inches.

Use the Z score formula: \frac{X-\mu}{\sigma}

Therefore

\frac{0.496-0.499}{0.002}\leq z\leq \frac{0.504-0.499}{0.002}

\frac{-0.003}{0.002}\leq z\leq \frac{0.005}{0.002}

-1.5\leq z\leq 2.5

Now use the standard normal table and determine the probability of that a ball bearing will be acceptable.

P(-1.5\leq z\leq 2.5)=0.9938-0.0668=0.9270

We need to find the percentage of the bearings produced will not be acceptable.

So subtract it from 1 as shown.

1-0.9270=0.073

Hence, 7.3% percentage of the bearings produced will not be acceptable.

3 0
4 years ago
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