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Nimfa-mama [501]
3 years ago
5

The rainfall in a certain area was 50% below normal last year if the rainfall last year measures 22 inches what is the normal ra

infall for that area
Mathematics
2 answers:
lapo4ka [179]3 years ago
7 0

Answer:

Norma rainfall for the area = 22*2 = 44 inches

Step-by-step explanation:

So because the rainfall last year was 50% below normal, this means that it was half of what is known as the normal rainfall (percentage total is equal to a 100%, so 50% is half of it).

So since you know that 22 inches is 50% of the normal value, that means a 100% of the normal value will be 22*2 = 44 inches.

marta [7]3 years ago
4 0

Answer: 44 inches

Step-by-step explanation:

22 x 2 = 44

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Answer:

Area = 112.1 m^2

Step-by-step Explanation:

Given:

∆WXY

m < X = 130°

WY = x = 31 mm

m < Y = 26°

Required:

Area of ∆WXY

Solution:

Find the length of XY using Law of Sines

\frac{w}{sin(W)} = \frac{x}{sin(X)}

X = 130°

x = WY = 31 mm

W = 180 - (130 + 26) = 24°

w = XY = ?

\frac{w}{sin(24)} = \frac{31}{sin(130)}

Multiply both sides by sin(24) to solve for x

\frac{w}{sin(24)}*sin(24) = \frac{31}{sin(130)}*sin(24)

w = \frac{31*sin(24)}{sin(130)}

w = 16.5 mm (approximated)

XY = w = 16.5 mm

Find the area of ∆WXY

area = \frac{1}{2}*w*x*sin(Y)

= \frac{1}{2}*16.5*31*sin(26)

= \frac{16.5*31*sin(26)}{2}

Area = 112.1 m^2 (to nearest tenth).

8 0
3 years ago
Diana's painting statue she has 7/8 of a gallon of paint remaining each student requires 1/16 of a gallon of paint how many stat
julsineya [31]
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4 0
3 years ago
Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a
Diano4ka-milaya [45]

Answer:

a) The standard error would be of 8.58 pounds.

b) The margin of error is 14.11 pounds.

c) The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds

Step-by-step explanation:

a.What is the standard error?

The standard error is

s = \frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. So

s = \frac{47}{\sqrt{30}} = 8.58

The standard error would be of 8.58 pounds.

b.What is the margin of error at 90% confidence?

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find the margin of error M as such

M = z*s = 1.645*8.58 = 14.11

The margin of error is 14.11 pounds.

c. Using my sample of 30, what would be the 90% confidence interval for the population mean?

Lower bound: Sample mean subtracted by the margin of error.

247 - 14.11 = 232.89 pounds

Upper bound

247 + 14.11 = 261.11 pounds

The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds

4 0
3 years ago
Based on the pattern, what are the next two terms of the sequence?<br> 7, 7/4 , 7/16 , 7/64, 7/256
IrinaVladis [17]

Answer:

7/1024 and 7/4096

Step-by-step explanation:

4×4=16

4×16=64

4×64=256

4×256=1024

4×1024=4096

hope this helps!

5 0
3 years ago
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