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3 years ago

12

The school's computer lab goes through 55 reams of printer paper every 77 weeks. find out how long 33 cases of printer paper is

likely to last (a case of paper holds 8 reams of paper).

1 answer:

aalyn [17]3 years ago

4 0

33 cases of printer paper is likely to last about 370 weeks

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8. A right cone has a volume of 8,579 m3 and a radius of 16 m. Find its altitude.

Monica [59]

Answer:

Option A is correct.

Step-by-step explanation:

The formula used for finding the volume of right cone is:

Volume of Right cone = (1/3)π.r².h

We need to find altitude i.e h

Volume of cone=V = 8579 m^3

Radius=r = 16m

Altitude =h =?

Putting values,

8579 = (1/3) * 3.14 * (16)^2*h

8579 = 1/3 * 3.14 * 256 *h

8579 = 267.95 * h

=> h = 8579/267.95

h = 32.0 m

So, Altitude of right cone is 32.0 m

Option A is correct.

4 0

3 years ago

A puppy weighs 1 pound. what does the puppy weigh after 4 weeks

Alona [7]

I think it would be around 7 lbs
(i could be wrong but i think this is about right)

8 0

2 years ago

Help me please I don't understand it.

Dennis_Churaev [7]

Answer:

Sample Space = {A, B, C, D, E, F}.

Sample space for choosing C to F = {C, D, E, F}.

Step-by-step explanation:

All six letters are included in the first set of possible outcomes.

Four letters (C to F) are included in the second set of possible outcomes.

4 0

3 years ago

Read 2 more answers

A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain

svetlana [45]

Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

$\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55$

The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

$LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8$

The probability of having between 4 and 8 customers choosing the oversize version is:

$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

$P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328$

7 0

3 years ago

A company that manufactures hair ribbons knows that the number of ribbons it can sell each week, x, is related to the price p pe

mel-nik [20]

Given:

The number of ribbons it can sell each week, x, is related to the price p per ribbon by the equation:

$x=1000-100p$

To find:

The selling price if the company wants the weekly revenue to be $1,600.

Solution:

We know that the revenue is the product of quantity and price.

$R=xp$

$R=(1000-100p)p$

$R=1000p-100p^2$

We need to find the value of p when the value of R is $1600.

$1600=1000p-100p^2$

$1600-1000p+100p^2=0$

$100(16-10p+p^2)=0$

Divide both sides by 100.

$p^2-10p+16=0$

Splitting the middle term, we get

$p^2-8p-2p+16=0$

$p(p-8)-2(p-8)=0$

$(p-8)(p-2)=0$

Using zero product property, we get

$p-8=0$ or $p-2=0$

$p=8$ or $p=2$

Therefore, the smaller value of p is $2 and the larger value of p is $8.

5 0

3 years ago