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Akimi4 [234]
4 years ago
11

Someone please help

Mathematics
1 answer:
julia-pushkina [17]4 years ago
3 0

Answer:

(a) So the top of the hill is 365 feet above sea level.

(b) Checkpoint 5 is 185 feet higher than checkpoint 2.

Step-by-step explanation:

<u>Solution for (a):</u>

<u />

Checkpoint 2 is -218 feet above sea level.

The top of a hill rises 583 feet above Checkpoint 2.

The altitude of the top of the hill = -218 + 583 = 365

So the top of the hill is 365 feet above sea level.

<u>Solution for (b):</u>

<u />

Checkpoint 2 is -218 feet above sea level.

Checkpoint 5 is -33 feet above sea level.

Checkpoint 5 is -33 - -218 = 185 feet higher than checkpoint 2.

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How many times is 3 in 43.999 than the value of the digit 3 in 42.103
Dennis_Churaev [7]

Answer:

1. 14.666 ¯ over the 3

2. 14.034 ¯ over the 3

Step-by-step explanation:

5 0
3 years ago
Help I can't fail! pleaseeeee
VLD [36.1K]

Answer:


Step-by-step explanation:

I believe it's C.33 because if 4×33=132 the rest are smaller and one of the is equal,

So in my opinion It's 33

5 0
3 years ago
a right triangle is removed from a rectangle to create the shaded region shown below. find the area of the shaded region. be sur
ser-zykov [4K]

Answer:

36

Step-by-step explanation:

Objective: Find missing dimensions of the right triangle using the then subtract that area of the triangle from the rectangle's area.

Using the definition of a rectangle, the left vertical side measure is 7 and right vertical side is 6. So the missing legs of the triangle is 4 and 3.

Apply triangle formula, base x height divided by 2.

\frac{4 \times 3}{2}  = 6

Area of rectangle is length x width so the area is

7 \times 6 = 42

Subtract 6 from 42.

42 - 6 = 36

3 0
3 years ago
What is 80% of 12 equal to
Varvara68 [4.7K]
10.4. You multiply .80 times 12 and get 9.6. Then you add 9.6 to 0.80 and get 10.4
8 0
3 years ago
The reading speed of second grade students in a large city is approximately normal, with a mean of 90 words per
kari74 [83]

Answer:

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

P(x⁻>96) =0.0359

Step-by-step explanation:

<em>Explanation</em>:-

<em>Given sample size 'n' =10</em>

<em>mean of the Population = 90 words per minute</em>

<em>standard deviation of the Population =10 wpm </em>

<em>we will use formula</em>

<em>                            </em>Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }<em></em>

<em>Let X⁻  = 96</em>

                           Z = \frac{96-90 }{\frac{10}{\sqrt{10} } }

                          Z =  1.898

<em>The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute</em>

<em></em>P(X^{-}>x^{-} ) = P(Z > z^{-} )<em></em>

<em>                    = 1- P( Z ≤z⁻)</em>

<em>                    = 1- P(Z<1.898)</em>

                   = 1-(0.5 +A(1.898)

                   = 0.5 - A(1.898)

                   = 0.5 -0.4641 (From Normal table)

                  = 0.0359

<u><em>Final answer</em></u>:-

The probability that a random sample of 10 second grade students from  

                = 0.0359

4 0
3 years ago
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