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3 years ago

15

The scale on a map is 1in. = 2 mi. On the map, Greg’s house is 4in. from Melissa’s house. What is the actual distance from Greg’

s house to Melissa’s house?

1 answer:

maria [59]3 years ago

6 0

The distance from Greg's house to Melissa's is 8 mi.

Step-by-step explanation:

Multiply 4 by 2.

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3 years ago

Read 2 more answers

1. Y varies inversely with x. If y = 9 when x = -5, find the value of x when y=-6.

Gelneren [198K]

Answer:

Step-by-step explanation:

1 ) Y∝ 1 / X

Y = k / X

Y = 9 , X = -5

Putting the values in the relation above

9 = k / - 5

k = -45

Y = - 45 / X

Put Y = - 6

- 6 = - 45 / X

X = - 45 / -6

= 7.5

2 )

Luke and Nora can peel 12 carrots in 5 minutes

in 1 minute Luke and Nora can peel 12 / 5 carrot

in 1 minute Luke alone can peel 12/ 11 carrot .

In 1 minute Nora alone will peel (12 / 5 - 12/ 11 ) carrot

(12 / 5 - 12 / 11 )

= 2.4 - 1.1

= 1.3 carrot

In 1 minute Nora alone will peel 1.3 carrot

1.3 carrot in 1 minute

8 carrot in 8 / 1.3 minute

= 6.15 minutes.

In 1 minute Nora alone will peel 8 carrot in 6.15 minutes working alone.

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3 years ago

a rectangular prism has base edge lengths of 6" and 8". The base edge lengths of a right rectangular prism are 8" and 6". The ba

FromTheMoon [43]

Answer:

Step-by-step explanation:

The heights of neither the rectangular prism nor the triangular prism are given, so we don't know the volume of either.

If h is the height of the rectangular prism, then its volume is

6×8×h = 48

if the height of the triangular prism is h/2, then its volume is

(1/2)×24×8×(h/2) = 48

So we know the volumes are the same -- but we don't know what that volume is.

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What is the area of the triangle?

liubo4ka [24]

Answer:

area of triangle : 1/2 x length x widht

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3 years ago

The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ric

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Let $X$ denote the number of minutes spent brushing his teeth, and let $\mu$ be the mean and $\sigma$ the standard deviation for this distribution.

$\mathbb P(X$

The z-score corresponding to this probability is approximately $z=-0.2533$ , which means

$\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1$

Next, (note the sign change)

$\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98$

The corresponding z-score is approximately $z=2.0538$ , so you have

$\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2$

Solving the two equations for $\mu$ and $\sigma$ , you'll find that the mean is approximately $\mu=1.1098$ and the standard deviation is approximately $\sigma=0.4334$ .

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3 years ago