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AlekseyPX
3 years ago
13

Solve and graph the inequality - x/4 - 6 > -8

Mathematics
1 answer:
QveST [7]3 years ago
7 0
-x/4-6>-8
Multiply both sides by 4

-x-24>-32
Move the constant to the right

-x>-32+24

-x>-8
Change the signs

x<8
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A coin bank that excepts only nickels and dimes contains $9.15. There are three more than twice as many nickels as there are dim
valentinak56 [21]

45 dimes and 93 nickels were in bank

<em><u>Solution:</u></em>

Let "n" be the number of nickels

Let "d" be the number of dimes

We know that,

value of 1 nickel = $ 0.05

value of 1 dime = $ 0.10

<em><u>Given that There are three more than twice as many nickels as there are dimes</u></em>

Number of nickels = 3 + 2(number of dimes)

n = 3 + 2d ---- eqn 1

<em><u>Also given that coin bank that excepts only nickels and dimes contains $9.15</u></em>

number of nickels x value of 1 nickel + number of dimes x value of 1 dime = 9.15

n \times 0.05 + d \times 0.10 = 9.15

0.05n + 0.10d = 9.15  ---- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Substitute eqn 1 in eqn 2

0.05(3 + 2d) + 0.10d = 9.15

0.15 + 0.1d + 0.10d = 9.15

0.2d = 9

<h3>d = 45</h3>

From eqn 1

n = 3 + 2(45)

n = 3 + 90 = 93

<h3>n = 93</h3>

Thus 45 dimes and 93 nickels were in bank

3 0
3 years ago
Scores from an exam are normally distributed with a mean of 85 and a standard deviation of 5. Suppose 188 students take the exam
charle [14.2K]

The mean is \mu=85 and standard deviation is \sigma=5. Then the variable X\sim N(85,5).

Use substitution Z=\dfrac{X-\mu}{\sigma}=\dfrac{X-85}{5}, then the variable Z\sim N(0,1).

The probability Pr(X>80) can be calculated in the following way:

for X=80, Z=\dfrac{80-85}{5} =-1 and Pr(X>80)=Pr(Z>-1). From the diagram Pr(Z>-1)=0.34+0.34+0.135+0.025=0.84.

Conclusion: if 188 students take the exam, then 188·0.84=157.92 (157) would receive a score above 80

4 0
3 years ago
Read 2 more answers
How do you solve x^2-4X-12=0
Pepsi [2]
                                         x²  -  4x  - 12  =  0

Factor the left side:      (x + 2) · (x - 6)  =  0

The equation is true if either factor is zero.

If  (x + 2) = 0  then    x = -2 .

If  (x - 6) = 0  then     x = 6 .
3 0
3 years ago
I AM IN A TEST AND I NEED THE ANSWER SOON!!! PLEASEE!!!<br><br> THANK YOU SO MUCH!!
mart [117]
Hello,

If Viviana has 2/5 of pasta to share with four friends, you should divide. First change 2/5 to a decimal, 0.4. Now divide 0.4 and 4.

Workspace:


2/5 = 0.4

4 / 0.4 -> 10

Workspace End

Correct Answer:

10

Hope this helps!!
Brainliest??
8 0
2 years ago
Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?
likoan [24]

\mathrm{Use\:the\:rational\:root\:theorem}

a_0=15,\:\quad a_n=2

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5

=\left(x+3\right)\left(2x^2-11x+5\right)

Factor: 2x^2-11x+5

2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)

=x\left(2x-1\right)-5\left(2x-1\right)

2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)

\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:}

thus zeros of f(x) is

x=-3,\:x=5,\:x=\frac{1}{2}

5 0
2 years ago
Read 2 more answers
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