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3 years ago

Helpppppppp please!!!!!!!!!!!

Mathematics

1 answer:

fenix001 [56]3 years ago

6 0

Answer:

a. (4, 13/6)

b. (4, -5/6)

c. (4, 5/6)

Step-by-step explanation:

sorry if it is not right.

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Which inequality is true?

Eduardwww [97]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The diameters of ball bearings are distributed normally. The mean diameter is 106 millimeters and the standard deviation is 4 mi

Assoli18 [71]

Answer:

Probability that the diameter of a selected bearing is greater than 111 millimeters is 0.1056.

Step-by-step explanation:

We are given that the diameters of ball bearings are distributed normally. The mean diameter is 106 millimeters and the standard deviation is 4 millimeters.

<em>Firstly, Let X = diameters of ball bearings</em>

The z score probability distribution for is given by;

Z = $\frac{ X - \mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean diameter = 106 millimeters

$\sigma$ = standard deviation = 4 millimeter

Probability that the diameter of a selected bearing is greater than 111 millimeters is given by = P(X > 111 millimeters)

P(X > 111) = P( $\frac{ X - \mu}{\sigma}$ > $\frac{111-106}{4}$ ) = P(Z > 1.25) = 1 - P(Z $\leq$ 1.25)

= 1 - 0.89435 = 0.1056

Therefore, probability that the diameter of a selected bearing is greater than 111 millimeters is 0.1056.

4 0

3 years ago

A camel had to walk a distance of 11,040 m. After 4 hours, he still had to walk 920 m. How far did the camel walk per hour?

sergey [27]

He walked 2,525 m per an hour.

11,040 - 920 = 10,100

10,100 / 4 = 2,525

Please mark brainliest.

7 0

3 years ago

What is the solution for this problem-2x+-6=8

viva [34]

Answer: X=7

hope that will help!

Step-by-step explanation:

4 0

3 years ago

1 If a = p^1/3-p^-1/3<br>prove that: a^3 + 3a = p - 1/p

alexandr402 [8]

Hello, please consider the following.

We know that

$a = p^{\frac{1}{3}}-p^{-\frac{1}{3}}\\\\=p^{\frac{1}{3}}-\dfrac{1}{p^{\frac{1}{3}}}$

And we can write that.

$(p-\dfrac{1}{p})^3=(p-\dfrac{1}{p})(p^2-2+\dfrac{1}{p^2})\\\\=p^3-2p+\dfrac{1}{p}-p+\dfrac{2}{p}-\dfrac{1}{p^3}\\\\=p^3-\dfrac{1}{p^3}-3(p-\dfrac{1}{p})$

It means that, by replacing p by $p^{1/3}$

$(p^{1/3}-\dfrac{1}{p^{1/3}})^3=p-\dfrac{1}{p}-3(p^{1/3}-\dfrac{1}{p^{1/3}})\\\\\\\text{ So }\\\\a^3=p-\dfrac{1}{p}-3a\\\\\boxed{ a^3+3a=p-\dfrac{1}{p} }$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0

3 years ago