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svetoff [14.1K]
2 years ago
5

Find the area of a circle with a circumference of 20 units squared

Mathematics
2 answers:
nlexa [21]2 years ago
6 0
31.83

remember area : pi radius squared
lara [203]2 years ago
4 0

Answer:

it might be 31.97 but i’m seriously not sure

Step-by-step explanation:

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A line with a slope of 9/7 passes through the point (7, 14).What is its equation in slope -intercept form?
ad-work [718]

Answer:

y=9/7x+5

Explanation:

We have the slope, and a point that is not an intercept. We can use Desmos graphing calculator to test y-intercepts.

6 0
2 years ago
Can someone give me the answer? I’m really confused
Damm [24]

1 pound of rice costs $3.45

1.2 pounds of rice costs x

Set up the proportion

1 / 1.2 = 3.45 / x               Cross Multiply

x = 1.2 * 3.45                   Combine

x = $4.14                          Answer

7 0
2 years ago
If each cube has edges 1.5 centimeters long, what is the volume of the prism outlined in blue?
aksik [14]

Answer:

91.125 cm3

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
-10 = y/11 - 13<br><br> solve for y
Flauer [41]

Answer:

10=y/11-13

We move all terms to the left:

-10-(y/11-13)=0

-y/11+13-10=0

We multiply all the terms by the denominator

-y+13*11-10*11=0

We add all the numbers together, and all the variables

-1y+33=0

We move all terms containing y to the left, all other terms to the right

-y=-33

y=-33/-1

y=+33

6 0
2 years ago
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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
2 years ago
Read 2 more answers
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