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kumpel [21]
3 years ago
10

Substitute -3x-8y=4 -2x+7y=15

Mathematics
1 answer:
Nikolay [14]3 years ago
3 0
-2x=15-7y
x=-7.5+3.5y

-3x-8y=4
-3(7.5+3.5y)-8y=4
-22.5-10.5y-8y=4
-22.5-18.5y=4
-18.5y=4+22.5
-18.5y=26.5
y=-53/35

the fraction might not be in lowest terms.
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Evelyn can run 3 miles in 40 minutes
hram777 [196]

Answer:

m= 4.5 miles

Step-by-step explanation:

Let the number of miles Evelyn can run be m and can be modeled by the equation below

m= kx

Where x Is the Minutes

When m= 3

x= 40

3= k40

k= 3/40

If x= 60

m=3/40(60)

m = (60*3)/40

m= 180/40

m= 4.5 miles

5 0
3 years ago
The set of points (1, 3), (3, 5), (4, 7) is collinear. ture or false
Mekhanik [1.2K]
(1,3)(3,5)
slope = (5-3) / (3-1) = 2/2 = 1

(3,5)(4,7)
slope = (7-5) / (4-3) = 2/1 = 2

False....they are not collinear because they have different slopes and do not lie on the same line
3 0
3 years ago
A cafeteria sells 6 times as many bottles of juice. if the cafeteria sells 750 bottles of water. how many bottles of juice did t
faltersainse [42]

Answer:

<h2>4,500 bottles of juice.</h2>

Step-by-step explanation:

<h3>Clue word: Times means multiply</h3><h3>If you multiply 750 x 6 you will get 4,500.</h3>
5 0
2 years ago
Reagan used 1,297 sq ft of wrapping paper to wrap gifts for everyone in her class. If each gift used the same amount of wrapping
seraphim [82]
She used 54.04 sq ft of wrapping paper
5 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
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