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3 years ago

HELP MEE HURYyyYYYYYYYY

Mathematics

1 answer:

Delicious77 [7]3 years ago

4 0

Answer:

45(1+2)

Step-by-step explanation:

Looking at you options, you can see that 45 is your Greatest common factor (GCF)

45 X 1 = 45

45 X 2 =90

45(1+2)

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Sonia's check at Sun Bistro is

Neporo4naja [7]

Answer:$3.672

rounded is $3.67

Step-by-step explanation:

5 0

3 years ago

Help me with 26,25,24

nexus9112 [7]

26: -90 / 5 = -18
25: -2 / 5 = -0.4
24: -5 / 5 = -1

4 0

3 years ago

Read 2 more answers

2 1/5 × 5/6 (its a fraction) PLSSS i need it now its a test

Soloha48 [4]

Answer:

$\frac{11}{6} \ \ or \ 1\frac{5}{6}$

Step-by-step explanation:

$2\frac{1}{5} \times \frac{5}{6} \\\\ = \frac{11}{5} \times \frac{5}{6} \\\\=\frac{11}{6}\\\\=1 \frac{5}{6}$

6 0

3 years ago

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A shopper bought a 12-pound bag of oranges for 18.75. What is the unit price per ounce?

noname [10]

Answer:

About $0.10 per ounce

Step-by-step explanation:

12 pound costs $18.75. Lets find the cost per pound first:

Cost Per Pound = $\frac{18.75}{12}=1.5625$

We know, there are 16 ounces in 1 pound.

We know 1 pound costs 1.5625

To find cost per ounce, we have to divide this by 16.

So,

Cost Per Ounce = $\frac{1.5625}{16}=0.097$

Rounded to nearest cent, that would be 10 cents per ounce

5 0

4 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

3 years ago