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3 years ago

Which statements are correct for all values of x?

1 answer:

8 0

A), C), D), and E are all true because if you were to solve them, both sides of the = sign would be the same number. For example, if I were to solve the first one (A), I would get 16x=16x

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Find the mean, variance, and standard deviation of the binomial distribution. (round to the

shusha [124]

Sixty eight and nine plus eleven equals 12

8 0

2 years ago

Go to a clothing store with

olga2289 [7]

Answer:

original price $75

discount $18.75

sale price $56.25

Step-by-step explanation:

I found a pair of dress pants for $75.

Since the discount is 25%, then the discount is

25% of $75 = 0.25 × $75 = $18.75

The sale price is the original price, $75, minus the discount, $18.75.

$75 - $18.75 = $56.25

5 0

2 years ago

A package of pencils cost $6.25 per set at an office supply store. Which equation can be used to determine the price, P, of n se

ladessa [460]

To solve, it would be 6.25n. That eliminates both 6n25 over P, P equals 6.25 over n, and n=6.25P, leaving P=6.25n as the answer.

5 0

3 years ago

Read 2 more answers

In the poportion 1/z =4/5/8 which number is equal to z in the proportion

Nady [450]

Answer:

case 1) $z=10$

case 2) $z=5/32$

Step-by-step explanation:

case 1) we know that

Using proportion

$\frac{1}{z}=\frac{(4/5)}{8}$

solve for z

$\frac{1}{z}=\frac{(4/5)}{8}\\ \\z(4/5)=8\\ \\z=8*5/4\\ \\z=10$

case 2) we know that

Using proportion

$\frac{1}{z}=\frac{4}{5/8}$

solve for z

$\frac{1}{z}=\frac{4}{5/8}\\ \\4z=5/8\\ \\z=5/32$

8 0

3 years ago

Tags are attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the e

ioda

Answer:

Step-by-step explanation:

Given that tags are attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the event that the right leg tag is lost. Suppose those two events are independent and A2 the event that the right leg tag is lost. Suppose those two events are independent and P(A1) = P(A2) = 0.3

a) the probability that at least one leg tag is lost

= $P(A_1 U A_2)\\= P(A_1) + P(A_2)-{(A_1 \bigcap A_2)\\=P(A_1) + P(A_2)-{(A_1 )P(A_2)$

(since A1 and A2 are independent)

=0.3+0.3-0.09

=0.51

b) the probability that exactly one tag is lost, given that at least one tag is lost.

= P(one leg lost)/P(atleast one leg lost)

= $P(atleast one leg lost)-P(Both lost)/0.51\\= \frac{0.51-0.09}{0.51} \\=14/17$

c) the probability that exactly one tag is lost, given that at most one tag is lost.

4 0

3 years ago