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algol [13]
3 years ago
5

95 more than the quotient of an unknown number and 75 is 96

Mathematics
1 answer:
zlopas [31]3 years ago
5 0
Translate this sentence into an equation:  (n/75) + 95 = 96
Subtracting 95 from both sides:                   n/75 = 1
Mult. both sides by 75 results in                     n = 75
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Consider 3 trials, each having the same probability of successes. Let X denote the total number of successes in these trials.
Mariulka [41]

Since each trial has the same probability of success, 

Let, <span><span><span>Xi</span>=1</span></span> if the <span><span>i<span>th</span></span></span> trial is a success (<span>0</span> otherwise). Then, <span><span>X=<span>∑3<span>i=1</span></span><span>Xi</span></span><span>X=<span>∑<span>i=1</span>3</span><span>Xi</span></span></span>, 

and <span><span>E[X]=E[<span>∑3<span>i=1</span></span><span>Xi</span>]=<span>∑3<span>i=1</span></span>E[<span>Xi</span>]=<span>∑3<span>i=1</span></span>p=3p=1.8</span><span>E[X]=E[<span>∑<span>i=1</span>3</span><span>Xi</span>]=<span>∑<span>i=1</span>3</span>E[<span>Xi</span>]=<span>∑<span>i=1</span>3</span>p=3p=1.8</span></span>

So, <span><span>p=0.6</span><span>p=0.6</span></span>, and <span><span>P{X=3}=<span>0.63</span></span><span>P{X=3}=<span>0.63</span></span></span>

I thought what I did was sound, but the textbook says the answer to (a) is <span>0.60.6</span> and (b) is <span>00</span>.

Their reasoning (for (a)) is as follows:

3 0
3 years ago
Make a the subject of relation x=√1/a square-c/d​
klasskru [66]

I hope I wrote the question rightly

4 0
3 years ago
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
What's the forecasted capital expenditure based on the information below? • Net PP&amp;E beginning of period: 15,000 • Net PP&am
SashulF [63]

The complete question is-

What’s the forecasted capital expenditure based on the information below?

Net PP&E beginning of period: 15,000

Net PP&E end of period: 17,500

Depreciation expenses: 2,400

Review Later

a) 2,500

b) 4,900

c) 100

d) -100

The forecasted capital expenditure related to the information below exists at 4,900.

Therefore, the correct answer is option b) 4,900.

<h3>What is Capital?</h3>

Capital exists directed as the lifeblood of any business. It exists the group of assets of the business that contains their financial value to create the production of goods and services.

For calculating the forecasted capital-

Net PP&E beginning of period = 15,000

Net PP&E end of period = 17,500

Depreciation expenses = 2,400

Forecasted capital expenditure = Net PP&E end of period + Depreciation expenses - Net PP&E beginning of the period

= 17500 + 2400 - 15,000

Forecasted capital expenditure = 4,900

Therefore, the correct answer is option b) 4,900.

To learn more about Capital, refer to:

brainly.com/question/16179329

#SPJ9

7 0
2 years ago
Solve using division or multiplication 7/2=14
professor190 [17]

Step-by-step explanation:

its false

7 0
3 years ago
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