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1 year ago

12

Sherry was in charge of distributing 25 food items

1 answer:

Valentin [98]1 year ago

6 0

Answer:

? i dont think you finished the sentence

Step-by-step explanation:

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Farmer Hanson is putting together fruit baskets. He has 240 apples and 150 pears. What is the largest number of baskets he can p

Tomtit [17]

Answer: 30 baskets.

<u>Step-by-step explanation:</u>

You need to find the Greatest Common Factor (GCF).

240 (apples) = 2 x 2 x 2 x 2 x 3 x 5

150 (pears) = 2 x 3 x 5 x 5

GCF (240, 150) = 2 x 3 x 5

= 30

You can make 30 baskets containing 240/30 = 8 apples and 150/30 = 5 pears.

4 0

1 year ago

A race car is driven by a professional driver at 99 . What is this speed in and ? 1 mile = 1.61 kilometers 1 hour = 60 minutes E

Lena [83]

Let us assume that the given speed is in miles per hour.

A race car is driven by a professional driver at 99 mph.

We have to find the speed in kilometer per hour and kilometer per minutes

As 1 mile = 1.61 km

So, 99 miles = $99\times1.61=159.39$ kmph

1 km per hour = 0.0166 km per minute

So, 159.39 kmph = $159.39\times0.0166=2.6458$ km per minute

Hence, The speed is equivalent to 159.39 kilometers /hour, or 2.645 kilometers/minutes .

7 0

1 year ago

Read 2 more answers

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

3. Standard deviation is ________. a mean of squared differences the square root of the variance unit-free the covariance 4. Sta

Ksenya-84 [330]

Answer:

3. Standard deviation is the square root of the variance.

4. Standard deviation is useful because it has the same units as the underlying data.

Step-by-step explanation:

3. In statistics, the dispersion in a given data with respect to its mean distribution can be determined or measured by standard deviation and variance. The standard deviation of a distribution can also be determined as the square root of variance.

4. Standard deviation is measured in the same units as that of the original data. Thus it has the same units as the underlying data.

5 0

1 year ago

Round to the nearest benchmark fraction 5/9

Stells [14]

1/2 because 5/9 is equivalent to 10/18. Half of 18 is 9 and 10 is close to 9 so the nearest benchmark fraction you should round to is 1/2. Hope this helps you!

5 0

2 years ago