The total weight of 25 books is 125 pounds. How many pounds would 70 books weigh?

Lin has a scale model of a modern train. The model is created at a scale of 1 to 48.

satela [25.4K]

Answer:A

1/48 = 102/ ???

1 × 102 = 102

48 × 102 = 4896 mm, or 4.896 m

B

1 inch = 25.4 mm

1/48 = 1 and one fourth in/54 in

54 ÷ 1 1/4 = 43

43 ≠ 48

The train can't drive on the tracks.

Step-by-step explanation:

3 0

3 years ago

3

adoni [48]

9514 1404 393

Answer:

(d) 2 11/12

Step-by-step explanation:

$\dfrac{2}{3}-\dfrac{5}{4}+\dfrac{7}{2}=\dfrac{2}{3}+\left(\dfrac{14}{4}-\dfrac{5}{4}\right)=\dfrac{8}{12}+\dfrac{27}{12}=\dfrac{35}{12}=\boxed{2\dfrac{11}{12}}$

7 0

3 years ago

A 6 foot person 15 feet from a streetlight casts a 15 foot shadow.Two similar triangles are formed.One triangle is formed by the

kipiarov [429]

What's the question?

6 0

3 years ago

Determine mu Subscript x overbar and sigma Subscript x overbar from the given parameters of the population and sample size. mu e

Mademuasel [1]

Answer:

$\bar x \sim N(\mu_{\bar x}=86, \sigma_{\bar x}=3)$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable who represents the variable of interest. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =86,\sigma =24)$

We select a sample of size n=64. That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The mean for the sample distirbution would be given by:

$\mu_{\bar x}=86$

And the deviation given by:

$\sigma_{\bar x} =\frac{24}{\sqrt{64}}=3$

And then the distribution for the sample mean is:

$\bar x \sim N(\mu_{\bar x}=86, \sigma_{\bar x}=3)$

7 0

3 years ago

A small family home in tucson arizona has a rooftop area of 1967 square feet, and it is possible to capture rain falling on abou

strojnjashka [21]

Answer:

Volume = 36.4 m^3 or 9616.6 gallon

weight = 18025.47 N or 80213.33 lbs

Step-by-step explanation:

The complete question is

A small family home in Tucson, Arizona, has a rooftop area of 1967 square feet, and it is possible to capture rain falling on about 56% of the roof. A typical annual rainfall is about 14 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water weigh when the tank was full (in N and in lbf)?

The area of the roof = 1967 ft^2

Area of water to be possibly captured = 56% of 1967 ft^2 = 0.56 x 1967 = 1101.52 ft^2

Rainfall depth = 14 inches = 14/12 ft = 1.167 ft

Volume of water = (area of water) x (depth of water) = 1101.52 ft^2 x 1.167 ft = 1285.47 ft^3

35.315 ft^3 = 1 m^3

1285.47 ft^3 = $x$ m^3

==> 1285.47/35.315 = 36.4 m^3

1 ft^3 = 7.481 gallon

1285.47 ft^3 = $x$ gallon

==> 1285.47 x 7.481 = 9616.6 gallon

Weight of water W = ρv

where ρ is the density of water = 62.4 lbs/ft^3

v is the volume of water = 1285.47 ft^3

substituting values,

W = 62.4 x 1285.47 = 80213.33 lbs

but,

1 lb = 4.45 N

therefore,

80213.33 lbs = 80213.33/4.45 = 18025.47 N

7 0

3 years ago