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bazaltina [42]
3 years ago
13

Anyone wanna help me please?!

Mathematics
1 answer:
photoshop1234 [79]3 years ago
4 0
<span>6:2=3
8:2=4

3 in by 4 in</span>
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Here are the city gas mileages for 13 different midsized cars in 2008. 16, 15, 22, 21, 24, 19, 20, 20, 21, 27, 18, 21, 48 What i
Vaselesa [24]

Answer:

Q1 --> 18.5

Step-by-step explanation:

4 0
3 years ago
HELP ME PLZZZZ!!!!!<br><br><br> It’s just 21 and 22
galina1969 [7]

Answer:

21) 12 cm

22) 5.9 cm

Step-by-step explanation:

21) The base is 10, and if the other two sides are congruent and the perimeter is 36, we can figure out with simple algebra that the sides are 13 cm long.

Half of 10 is 5, so we can use the pythagorean theorem.

5^2+x^2=13^2

Rearranging the variables we have 169-25=x^2

144=x^2

x can be plus or minus 12, but since negative length is impossible we find that x is positive 12 cm.

22) We want to use sine, because we have opposite and hypotenuse. A simple and easy way to memorize this is the SohCahToa method. If we have opposite (O) and hypotenuse (H) we have OH. Soh has the letters O and H, and the S means we should use sine.

sine 36=a/10

Plug this into a calculator or desmos scientific calculator to get a=5.9

7 0
3 years ago
The product of nine and the total of r and eight
yuradex [85]

Answer:

9(r + 8) = 9r +72

Step-by-step explanation:

9(r + 8)

9r + 72

Hope this helps!

7 0
2 years ago
A school sells pencils to student for 25 cents each.If a student has $5, how many pencils can the student buy?​
Ne4ueva [31]
The student can buy 20

Explanation

$5/$0.25 is 20
4 0
4 years ago
Read 2 more answers
the diameters of Douglas firs grown at a Christmas tree farm are normally distributed with a mean of 4 inches and a standard dev
aksik [14]

Answer:

Proportion of the trees will have diameters between 2 and 6 inches = 0.8164

Step-by-step explanation:

Given -

Mean (\nu )  = 4

Standard deviation (\sigma  ) = 1.5

Let X be the diameter of tree

proportion of the trees will have diameters between 2 and 6 inches =

P(2<  X<  6)   =  P(\frac{2 - 4 }{1.5}< \frac{X - \nu }{\sigma}<  \frac{6 - 4 }{1.5})

                         = P(\frac{-2 }{1.5}< Z<  \frac{2 }{1.5})     Put  [Z = \frac{X - \nu }{\sigma}]

                         =  P(-1.33< Z<  1.33)

                          = (Z<  1.33) - (Z<  -1.33)

                          = .9082 - .0918

                           = 0.8164

6 0
4 years ago
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