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adoni [48]
2 years ago
9

A car accelerates at 3.00 m/s2. The car has a mass of 1354kg. What net force will produce the acceleration?

Mathematics
1 answer:
otez555 [7]2 years ago
7 0

Answer:

4062N

Step-by-step explanation:

Using Newton’s Second Law;

F = ma

F = (1354)(3)

F = 4062N

Hope this helps! :D

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Alicia drove 265 miles in 5 hours. What is the average rate that she traveled?
Bumek [7]
265/5 = 53/1

53 miles per hour. 

*I hope this helped!
5 0
3 years ago
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Which ordered pair is a solution of the equation?<br> y = 3x+5
DedPeter [7]

Answer:

Step-by-step explanation:

Y=1/3y+-5/3

7 0
3 years ago
A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer
pshichka [43]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

Z_{o.o5}= -1.648

You reverse the standardization using the formula Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 7.72ounces

The lower control point will be 7.72 ounces.

The upper value will be:

Z_{0.95}= 1.648

1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

X[bar]= 8.28ounces

The upper control point will be 8.82 ounces.

b. Now μ= 7.6, considering the control limits of a.

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))

P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242

There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

5 0
3 years ago
Read 2 more answers
Find p(-4) and p(2) for the function p(x)=11x^5-11x^4-5x^2+15x-8
egoroff_w [7]

hi,

you must replace x by the number between parenthese.

I show you with the first one and let you do the second one

p(x) = 11x^5 -11x^4 - 5x^2 +15x-8

p(-4)  =   11 (-4)^5 - 11 (-4)^4 -5(-4)² +15(-4) -8

p(-4) =     11  ( -1024- 256) - 5*16 -60-8

p(-4) =   11 ( -1280) -80-60-8

p(-4)   =    - 14080 - 148

p(-4) =   - 14 228

7 0
3 years ago
I also need help figuring this out - <br><br> 2x^2 -7x - 15
Alexus [3.1K]
x = ±√ 7.500 = ± 2.73861
3 0
3 years ago
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