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alexgriva [62]
2 years ago
10

PLEASE HELP!!! ILL GIVE THE BRIANLIEST TO TO WHOEVER ANSWERS CORRECTLY!!!!!!!

Mathematics
1 answer:
Mariulka [41]2 years ago
6 0

Answer:

B

Step-by-step explanation:

The range is the largest number minus the smallest number.

Hope this helps!

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12=y-3y what's the answer
nevsk [136]
12=y-3y  Combine like terms
12=-2y  Divide
y=-6
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3 years ago
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deff fn [24]

Answer:

b = 12

Step-by-step explanation:

b/3 + 4 = 8

subtract 4 on both sides

b/3 = 4

multiply 3 with 4

b = 12

4 0
2 years ago
Suppose Casey Title Company normally charges $300 for services related to selling a house. As part of a summer special, Casey of
g100num [7]

Given :

Casey's company charges , C = $300.

In summer special, Casey offers customers a trade discount of 25%.

To Find :

Charges during summer.

Solution :

Let, x is the amount of charges they take during summer.

Casey offers customers a trade discount of 25%.

x =300(1-\dfrac{25}{100})\\\\x=300(1 - 0.25)\\\\x=300(0.75)\\\\x=\$225

Therefore, amount of charges during summers are $225.

Hence, this is the required solution.

8 0
3 years ago
What is the value of c so that y=x^2+15x+c is a perfect square.
Verizon [17]
Divide 15 by 2, then square the amount.

(15/2)^2 = 225/4
7 0
3 years ago
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A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers
Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b)  With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57

Using normal distribution table, z_{0.005 is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

e= z_{0.005}\frac{\sigma}{\sqrt{n} }  = 2.57*\frac{3900}{\sqrt{100} } =1002.3

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) =  (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0
3 years ago
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