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4 years ago

What is the rate of $120 for 15 books written as a unit rate? thank u

Mathematics

1 answer:

Vikki [24]4 years ago

4 0

Answer:

1 book for 8 1/8

Step-by-step explanation:

120/15 simplify by dividing by 8 and get 1/8

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Katie solved the equation by completing the square. X2+12x−7=0 which equation shows one of the steps katie could have taken to c

sdas [7]

solution:

$x^{2}+12x-7=0\\add 7 to both side of the equation\\x^{2}-12x=7\\to creat a trinomial square on the left side of the equation.\\find a value that is equal to the square of b and the cofficent of x.\\(\frac{b}{2})^2=(-6)^2\\add the term to each side of the equation\\x^2-12x+(-6)^2=7+(-6)\\x^2-12x+36=43\\(x-6)^2=43\\solve the equation for x.\\$ $take the square root of each side of the equation to the set up the solution x,\\(x-6)^2.\frac{1}{2}=\pm \sqrt{43}\\x-6=\pm \sqrt{43}\\x=6\pm \sqrt{43}$

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3 years ago

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Pam and her sister have a pattern of saving money. Pam saved 40 dollars and her brother saved 100 dollars.

Keith_Richards [23]

Answer:

true

Step-by-step explanation:

Pam saved 40

her bro saves 100

let us take this as ration 40 is to 100 which can also be written in fractional for like 40/100 now we simplify it how by dividing both by 20 so (40/20)/(100/20) which will give 2/5 which can be written as 2 is to 5 so this is true

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3 years ago

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The set in order from least to greatest

bonufazy [111]

Answer:

Hello there user! Could you please provide the set to order? Without given information, no one is able to help you out with your question.

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3 years ago

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What is the greatest common factor of 4k, 18k^4 and 12

vekshin1

Greatest Common Factor: 2

6 0

3 years ago

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Suppose a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101

stepan [7]

Answer:

101, 3, 0.025, 0.7416

Step-by-step explanation:

Given that a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.

As per central limit theorem we have

a) Mean of sample mean = $E(\bar x) =\\ \mu =101$

Std deviation of sample mean = $\frac{\sigma}{\sqrt{n} } =3$

Mean = 101

Std dev =3.00

b) $P(\bar x >107) = P(Z>\frac{107-101}{3} )\\= P(Z>2) = 0.025$

c) the probability that the sample mean deviates from the population mean μ = 101 by no more than 4.

= $P(|\bar x-101|) \leq 4\\= P(|z|\leq 1.13)\\= 2(0.3708)\\=0.7416$

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3 years ago