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Alenkasestr [34]
2 years ago
10

An art teacher needs to buy at least 60 brushes for her class. The brushes are sold in packs of 8.

Mathematics
1 answer:
kari74 [83]2 years ago
4 0

Answer:

The art teacher needs to buy at least 8 packs of 8 brushes.

Step-by-step explanation:

Every pack has 8 brushes. She needs AT LEAST 60 so she cannot have less than 60 brushes but can have more than 60. 8 x 8 =64 so she would have 4 bushes left over. She needed to buy 8 packs so she can have at least 60.

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Sammie bought just enough fencing to border either a rectangular plot or a square plot, as shown. The perimeters of the plots ar
Gekata [30.6K]
Perimeter of square : p = 4a.....
p = 4(x + 2)
p = 4x + 8

perimeter of rectangle : p = 2(L + W)
p = 2(3x + 2 + x - 1)
p = 2(4x + 1)
p = 8x + 2

so if the perimeters are te same, lets set them equal to each other and solve for x

4x + 8 = 8x + 2
8 - 2 = 8x - 4x
6 = 4x
6/4 = x
1.5 = x

the square : p = 4x + 8.....p = 4(1.5) + 8.....p = 14 meters
the rectangle : p = 8x + 2....p = 8(1.5) + 2.....p = 14 meters

so she bought 14 meters of fencing <==
8 0
2 years ago
Ari mixed 2 1/4 cups of red grapes with 1/2 cups of green grapes. He then divided the grapes into bags with 3/4 cup of mixed gra
Oxana [17]

To start, our equation will look like this:

(2 1/4 + 1/2)/ 3/4 = b

To make it easier convert the 2 numbers inside the parentheses into improper fractions, and then multiply them by their least common denominator to add.

(9/4 + 1/2)

(18/8 + 4/8)

(22/8)/ 3/4 = b

Now, to divide 22/8 cups of grapes into bags with 3/4 cup of grapes, multiply 22/8 by 4/3.

(22/8)(4/3)=b

3 2/3 = b

So, if Ari has a bag with 2 1/4 cups of red grapes and 1/2 cup of green grapes, and he divides it into bags with 3/4 cup each, he will have about 3 full bags.

6 0
2 years ago
Read 2 more answers
You drop a seashell into the ocean from a height of 40 feet. Write an equation that models the height h in feet) of the seashell
ANTONII [103]

Answer:

t=1.58 seconds

Explanation:

The equation which models the height <em>h</em><em> </em>of the seashell above water after t seconds

5 0
2 years ago
Someone please help me in this question it is kind of confusing it is a question under pre calculus in first year calculus i wil
masya89 [10]
Neither P, nor A are on the sketch 
I guess P is the upper right corner of the rectangle 
and A=(0,1) 

P belongs to the line going through (1,0) and B(0,y) 
<span>but we don't know the y-coordinate of B </span>

<span>the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 </span>
now look at the right triangle BOA 
<span>his hypotenuse is AB=sqrt2 and the <span>the kathete</span> OA is 1 </span>
so y²+1²=(sqrt2)² ... y²+1=2 ... y²=1.. y=1 
so the coordinates of B are (0,1) 

the line going through (1,0) and (0,1) is L(x)=-x+1 

P belongs to this line, so the coordinates of P are P(x,-x+1) (0<x<1) 

b) so if that's P, the height of the rectangle is -x+1 and the width=2x 
<span>so its area A(x)=2x*(-x+1)= -2x²+2x 

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

</span>
5 0
2 years ago
Bones Brothers &amp; Associates prepare individual tax returns. Over prior years, Bones Brothers have maintained careful records
madreJ [45]

Answer:

For this case we have the following info related to the time to prepare a return

\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. 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}})

And the standard deviation would be:

\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2

And the best answer would be

b. 2 minutes

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

For this case we have the following info related to the time to prepare a return

\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. 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}})

And the standard deviation would be:

\sigma_{\bar X} =\frac{14}{\sqrt{49}}= 2

And the best answer would be

b. 2 minutes

3 0
2 years ago
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