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3 years ago

Jenny scored a 23/28 on her math test. Tess scored a 81 percent. Who

Mathematics

2 answers:

ivanzaharov [21]3 years ago

6 0

Answer:

tess

Step-by-step explanation:

Aleks04 [339]3 years ago

6 0

Answer:

Jenny scored higher.

Step-by-step explanation:

You can figure out easy percentages by dividing the smaller number by the larger number, this will give you the said percent.

Jenny scored a 82% on her test while Tess scored an 81%

Hope this helps!

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I know it small but can someone help me. It says The table below shows customers' favorite ice cream flavors at a popular ice cr

IgorLugansk [536]

Answer:

<h2><u>Graph A</u></h2>

Step-by-step explanation:

We know it must be graph A because when finding the most popularly bought item, the percentages only can go to 100%

So graphs C and D are eliminated.

Also since it never said any flavors were unknown, all the percents have to add up to 100%

So graph B is eliminated.

Only graph A follows the requirements

7 0

2 years ago

Combine the like terms to create an equivalent expression: <br> 2s-( -4s)= ?

Ghella [55]

Answer:

Step-by-step explanation:

1. 2s-(-4s)

2. 2s+4s

3. 6s

6 0

3 years ago

iris [78.8K]

Who bout to do all this

8 0

3 years ago

The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr

aniked [119]

Answer:

(a) $E(X) = 0.383$

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) $P(x = 4) = 0.000169$

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$E(X) = \frac{n \times r}{N}$

$E(X) = \frac{10 \times 8}{209}$

$E(X) = \frac{80}{209}$

$E(X) = 0.383$

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}}$

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}}$

$P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}}$

$$ P(x = 4) = \frac{70 \times 84944276340}{35216131179263320}$

$P(x = 4) = 0.000169$

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0

2 years ago

at the dog show there are 4 times as many boxers as spaniel if there is a total of 30 dogs. how many dogs are spaniel

alexgriva [62]

4x+x = 30

5x = 30

x = 30/5 =6

6*4 = 24 boxers

and 6 spaniels

5 0

3 years ago