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Mila [183]
3 years ago
10

Help me with these2 question plz and I will press the thanks button

Mathematics
1 answer:
den301095 [7]3 years ago
6 0
7) 62 - 9.817
= 52 . 183

8) 35.1 + 4.89
= 39 . 99


Hope I can help you :)

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I need help people what is 215÷ 9
Dominik [7]

215/9 = 23.888

or

23 with a remainder of 8

I hope I helped


6 0
3 years ago
Read 2 more answers
HELP!!! 10 points marked brainliest!!!! stuck
nexus9112 [7]
Answer: 43
Explanation: All triangles add up to 180. We can subtract angle Z to get 86. Since X and Y are equal, we can divide 86 by 2.
3 0
3 years ago
Need help please assist me find the area of a regular hexagon
garri49 [273]

Answer:

The area of the regular hexagon is 166.3\ units^{2}

Step-by-step explanation:

we know that

The area of a regular hexagon can be divided into 6 equilateral triangles

so

step 1

Find the area of one equilateral triangle

A=\frac{1}{2}(b)(h)

we have

b=r=8\ units

h=4\sqrt{3}\ units ----> is the apothem

substitute

A=\frac{1}{2}(8)(4\sqrt{3})

A=16\sqrt{3}\ units^{2}

step 2

Find the area of 6 equilateral triangles

A=(6)16\sqrt{3}=96\sqrt{3}=166.3\ units^{2}

7 0
3 years ago
What is the volume of a box that will hold exactly 567of these cubes with 1.3 inch side
Bas_tet [7]

Answer:

21 inch cube

Step-by-step explanation:

(1/3)(1/3)(1/3)=1/27 is the volume of the cube

 the volume of the box : 567*(1/27)=567/27=21

5 0
3 years ago
A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l
tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

<u><em>Step(ii):-</em></u>

<u><em>The Empirical rule</em></u>

         Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

        Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

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

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

3 0
3 years ago
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