How do you determine which place value to use when estimating products?

Find the least common multiple (LCM) of 54 and 72. A) 9 B) 27 C) 108 D) 216

True [87]

The lowest common multiple is 216 (D)

4 0

3 years ago

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Write A story that could represent This problem 2/6 x 3/4

stellarik [79]

Answer:

Bob at the store

Step-by-step explanation:

Bob is at a store and wants to buy a banana, last time he checked it costed 2/6 of a dollar. Now it costs 3/4 times more.

3 0

2 years ago

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Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation o

Vikki [24]

Answer:

(a) The percent of her laps that are completed in less than 130 seconds is 55%.

(b) The fastest 3% of her laps are under 125.42 seconds.

(c) The middle 80% of her laps are from 126.80 seconds to 132.63 seconds.

Step-by-step explanation:

The random variable X is defined as the number of seconds for a randomly selected lap.

The random variable X is normally distributed with mean, μ = 129.71 seconds and standard deviation, σ = 2.28 seconds.

Thus, $X\sim N(129.71,\ 2.28^{2})$ .

(a)

Compute the probability that a lap is completes in less than 130 seconds as follows:

$P(X$

$=P(Z$

The percentage is, 0.55 × 100 = 55%.

Thus, the percent of her laps that are completed in less than 130 seconds is 55%.

(b)

Let x represents the 3rd percentile.

That is, P (X < x) = 0.03.

⇒ P (Z < z) = 0.03

The value of z for the above probability is:

z = -1.88

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\-1.88=\frac{x-129.71}{2.28}\\x=129.71-(1.88\times 2.28)\\x=125.4236\\x\approx 125.42$

Thus, the fastest 3% of her laps are under 125.42 seconds.

(c)

Let x₁ and x₂ be the values between which the middle 80% of the distribution lie.

That is,

$P(x_{1}$

The value of z for the above probability is:

z = 1.28

Compute the values of x₁ and x₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-1.28=\frac{x_{1}-129.71}{2.28}\\x_{1}=129.71-(1.28\times 2.28)\\x=126.7916\\x\approx 126.80$ $z=\frac{x_{2}-\mu}{\sigma}\\1.28=\frac{x_{2}-129.71}{2.28}\\x_{2}=129.71+(1.28\times 2.28)\\x=132.6284\\x\approx 132.63$

Thus, the middle 80% of her laps are from 126.80 seconds to 132.63 seconds.

5 0

3 years ago

The lines below are perpendicular. If the green line has a slope of -2/3, what is the slope of the red line? Enter your answer a

11111nata11111 [884]

Let the red line be x
_2 by 3 _ x

3 0

3 years ago

The expression 148 – 24d gives the number of pizzas a store has in stock, where d is the number of days. How many pizzas will th

Feliz [49]

Explanation

A function is defined as a relation between a set of inputs having one output each,in this case our independent variable is the number of days( d) and the dependent variable would be the number of pizzas

so

Step 1

let the function

$\begin{gathered} P=148-24d \\ where\text{ P is the number of pizzas } \\ and\text{ d is the number of days} \end{gathered}$

therefore, to know the number of pizzas are in the stock after 5 days we need to evaluate the equation for d=3

so

$\begin{gathered} P=148-24d \\ let\text{ d=5 and evaluate} \\ P=148-24(5) \\ P=148-120=28 \\ P=28 \end{gathered}$

so, after five day there willbe 28 pizzas in stock

I hope this helpsyou

6 0

1 year ago