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3 years ago

15

Tim needs 18 pens. He can buy them in packages of 6, 9, 12.he will buy only one type of package. Which packages could tim buy. W

rite two different ways that to. Could buy exactly 18 pens

1 answer:

Mariulka [41]3 years ago

5 0

Two packages of 9.
• 9+9=18

OR

One package of 12 and one package of 6.
•12+6=18

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A rectangular box has faces with areas of 12,15, and 20 square units. What is the volume of the box?

miv72 [106K]

First find the side lengths of the box by factoring each area:

12: 1, 2, 3, 4, 6, 12
15: 1, 3, 5, 15
20: 1, 2, 4, 5, 10, 20

Find 3 numbers that both have 2 sides have in common and can multiply to make the areas of the faces:

12: 3 * 4
15: 3 * 5
20: 4 * 5

Sides: 3, 4, and 5

Then multiply to get the volume:

3 * 4 * 5 = 60 units^2

6 0

3 years ago

Farmer Jack needs 1,800 square feet of garden space to have enough corn for a year. His garden space is currently 10 5?6 feet by

sergey [27]

Answer:

$1,605\ ft^{2}$

Step-by-step explanation:

step 1

Find the area of the garden space

First convert mixed number to an improper fraction

$10\frac{5}{6}\ ft=\frac{10*6+5}{6}=\frac{65}{6}\ ft$

$A=\frac{65}{6}*18=195\ ft^{2}$

step 2

Find how much more land Farmer Jack needs to plant the rest of his corn

$1,800\ ft^{2}-195\ ft^{2}=1,605\ ft^{2}$

4 0

3 years ago

A student determined that the area of the segment of c shown above is Asegment = 137.71 ft2. The student's work is shown below.

ivolga24 [154]

Answer:

Option D. The student did not use the correct formula to calculate the area of the segment

Step-by-step explanation:

step 1

Find the area of the isosceles triangle

Applying the law of sines

$A=\frac{1}{2}(12^{2})sin(60\°)=62.35\ ft^{2}$

step 2

Find the area of the sector

The area of the sector is 1/6 of the area of the circle

so

$A=\pi r^{2}/6$

substitute the value

$A=(3.14)(12)^{2}/6=75.36\ ft^{2}$

step 3

Find the area of the segment

The area of the segment is equal to the area of sector minus the area of triangle

$A=75.36\ ft^{2}-62.35\ ft^{2}=13.01\ ft^{2}$

therefore

The student did not use the correct formula to calculate the area of the segment

4 0

3 years ago

A company compiles data on a variety of issues in education. In 2004 the company reported that the national college freshman-to

nasty-shy [4]

Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

And we have :

$\mu_p = 0.66$

$\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212$

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

And we have :

$\mu_p = 0.66$

$\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212$

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

8 0

3 years ago

SOLVE 3 AND 4 I WILL GIVE BRAINLIST!!

ZanzabumX [31]

Answer:

3. C≈31.42

4.C≈94.25

Step-by-step explanation:

10 ÷ 2 =5

C=2πr

c=2 (3.14) 5

c= 2(3.14)15

7 0

3 years ago