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3 years ago

5

Mr. Burdzy has a bowl of M&Ms. His least favorite color of M&M's is brown, so there are a lot of brown M&Ms left in

the bowl. How many brown M&M's are left in a bowl of 200 M&Ms if 60% of the bowl is brown

2 answers:

AlladinOne [14]3 years ago

6 0

Answer:

120 brown m&ms

Step-by-step explanation:

makvit [3.9K]3 years ago

5 0

Answer:

There are one hundred and twenty brown M&M's

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Can someone help me solve this please! I need to find what N equals!

Flura [38]

[ (12 - n)/7 ] = -1

12- n = -7

-n = -19

n = 19

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3 years ago

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A math professor notices that scores from a recent exam are normally distributed with a mean of 61 and a standard deviation of 8

Alexeev081 [22]

Answer:

a) 25% of the students exam scores fall below 55.6.

b) The minimum score for an A is 84.68.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 61 and a standard deviation of 8.

This means that $\mu = 61, \sigma = 8$

(a) What score do 25% of the students exam scores fall below?

Below the 25th percentile, which is X when Z has a p-value of 0.25, that is, X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 61}{8}$

$X - 61 = -0.675*8$

$X = 55.6$

25% of the students exam scores fall below 55.6.

(b) Suppose the professor decides to grade on a curve. If the professor wants 0.15% of the students to get an A, what is the minimum score for an A?

This is the 100 - 0.15 = 99.85th percentile, which is X when Z has a p-value of 0.9985. So X when Z = 2.96.

$Z = \frac{X - \mu}{\sigma}$

$2.96 = \frac{X - 61}{8}$

$X - 61 = 2.96*8$

$X = 84.68$

The minimum score for an A is 84.68.

8 0

3 years ago

Complete the steps to add 3 2/5 +1 3/10 . 1. Write mixed numbers as improper fractions. 2. Write equivalent fractions with a com

ira [324]

Answer:

47/10 = 4 7/10

Step-by-step explanation:

3 2/5 + 1 3/10

1. Write mixed numbers as improper fractions:

17/5 + 13/10

2. Write equivalent fractions with a common denominator:

34/10 + 13/10

3. Add the numerators over the common denominator.

47/10

4. Simplify.

4 7/10

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4 years ago

Read 2 more answers

A classroom has 12 students with brown hair, 8 students with black hair, and 5 students with blonde hair. If three students are

aksik [14]

I am going to have to go with, C, because the other answers are Invalid.

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3 years ago

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$

d) $t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)$

$N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h$

6 0

3 years ago