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zlopas [31]
3 years ago
9

What is the area of this triangle m2 19 m 22m

Mathematics
1 answer:
alexandr402 [8]3 years ago
4 0
Uhhh i don’t really know good luck figuring it out!!
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If edging cost $2.32 per 12-inch stone, and you want a double layer of edging around your flower bed that is 6 yards by 1 yard.
aksik [14]
12 inches = 1foot
1yard = 3feet
6yards = 18feet
Rerange the question:
If edging cost $2.32 per 1foot stone, and you want a double layer of edging around your flower bed that is 18feet by 3feet. How much will edging you flower bed cost?
Perimter:
2(18+3)=42
And since you want to double it; double the perimeter too.
42(2)=84

2.32(84)=194.88
7 0
3 years ago
Read 2 more answers
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
Solve for x.<br><br> 8x^2−3=2
sergeinik [125]

8x^2=2+3

8x^2=5

x^2=5/8

x= sqrt(5/8)

5 0
2 years ago
Which element of variation would not be affected by adding the data value 15 to the data set {3, 5, 6, 8, 9, 10, 13, 14}? A. the
Gwar [14]
B. The median of the data set

**Update; A. The lower quartile**
See other answer for explanation. 
8 0
3 years ago
Read 2 more answers
Use the premises and conclusion to answer the question.
Free_Kalibri [48]
I think it’s C, correct me if I’m mistaken.
5 0
2 years ago
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