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2 years ago

Multiply. Write your answer as a fraction in simplest form. 5/2*3/10

1 answer:

7 0

(5/2)(3/10)=(5*3)/(2*10)=15/20=3/4

Or

(5/2)(3/10)= (5)(1/2)(3)(1/10)=(5)(3)(1/2)(1/10) =(5*3)/(2*10)=15/20=3/4

Or

(5/2)(3/10)=(5)(0.5)(3)(0.1)=(2.5)(3)(0.1)=(7.5)(0.1)=.75=3/4

All of these work, but the first one is simplest

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Which of the following does an appropriately cautious reading of a receipt entail? I. Making sure that the items charged are the

Black_prince [1.1K]

An appropriate cautious reading of a receipt entail making sure that the

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When an item is bought, a receipt is usually issued which confirms that

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2 years ago

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Ede4ka [16]

Answer: b is the answer

Step-by-step explanation: have a nice day

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2 years ago

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

e-lub [12.9K]

Answer with Step-by-step explanation:

Since the given event is binary we can use Bernoulli's probability to sove the problem

Thus for an event 'E' with probability of success 'p' the probability that the event occurs 'r' times in 'n' trails is given by

$P(E)=\frac{n!}{(n-r)!\cdot r!}\cdot p^{r}\cdot (1-p)^{n-r}$

Part a)

For part a n = 11 , r =9, p = 0.75

Applying values we get

$P(E)=\frac{11!}{(11-9)!\cdot 9!}\cdot (0.75)^{9}\cdot (1-0.75)^{11-9}\\\\\therefore P(E)=0.2581$

Part b)

For part b n = 20 , r = 16 , p=0.75

Applying values we get

$P(E)=\frac{20!}{(20-16)!\cdot 16!}\cdot (0.75)^{16}\cdot (1-0.75)^{20-16}\\\\\therefore P(E)=0.1896$

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2 years ago

Dan has twice as many stickers as Cam does. If Dan gives Cam 15 of his stickers he will still have 22 more than Cam. How many st

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You can solve this question by setting up an equation. Dan has twice as many stickers as Cam does. We can call Dan 2x, and Cam x. So the equation would be:
2x-15=x+22.
Then solve this equation:
Add 15 on both sides
2x=x+37
Subtract x on both sides
2x-x=37
X=37.
Therefore Cam has 37 stickers, and Dan has 74 stickers.

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2 years ago

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Sati [7]

Answer:

The answer is A, ab<cd.

Step-by-step explanation:

If a, b, c, d are positive, and if a b < c d , then Ab>cd is always true

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2 years ago