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3 years ago

The average of 5 numbers is 80. If one number is removed, the average become 50. Find the removed number?

1 answer:

8 0

X / 5 = 80
x = 80 * 5
x = 400.......so the total of the numbers added was 400

(400 - x) / 4 = 50
400 - x = 50 * 4
400 - x = 200
-x = 200 - 400
-x = -200
x = 200 <== the removed number

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Olenka [21]

Answer:

$\frac{(9-x)}{3x}$

Step-by-step explanation:

we have

$\frac{\frac{2}{x-3}-\frac{3}{x}}{\frac{3}{x-3}}$

step 1

Solve the numerator of the quotient

$\frac{2}{x-3}-\frac{3}{x}=\frac{2x-3(x-3)}{(x-3)x}=\frac{2x-3x+9}{(x-3)x}=\frac{9-x}{(x-3)x}$

step 2

substitute in the original expression

$\frac{\frac{9-x}{(x-3)x}}{\frac{3}{x-3}}$

$\frac{(x-3)(9-x)}{3x(x-3)}$

step 3

simplify

$\frac{(9-x)}{3x}$

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3 years ago

4 out of 9 dogs weigh less than 20 pounds. What is the decimal equivalent for the number of dogs weighing under 20 pounds?

iVinArrow [24]

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6 0

3 years ago

Three more than the product of two and a number x

Nonamiya [84]

$2x+3$

You are taking two times a number, which is defined as x. That is equivalent to 2x. Then, you are adding three more to that product.

3 0

3 years ago

Read 2 more answers

What is the 11th term of the geometric sequence 3,6,12,24,…?

Mamont248 [21]

Answer:

The 11th term is 3072.

Step-by-step explanation:

First, find the common ratio and the first term. The common ratio is 6/3 or 2. The first term is 3.

Use the direct formula, xi=a*r^i-1, to find the eleventh term. a is first term, and i is the index or the number of the term.

Substitute: x11=3*2^(11-1)

Or x11=3*2^10

So x11 is 3072

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3 years ago

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following p

aev [14]

Answer:

P(X < 3) = 0.7443

Step-by-step explanation:

We are given that the random variable X has a binomial distribution with the given probability of obtaining a success. Also, given n = 6, p = 0.3.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 6

r = number of success = less than 3

p = probability of success which in our question is 0.3.

LET X = a random variable

So, it means X ~ $Binom(n=6, p=0.3)$

Now, Probability that X is less than 3 = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{6}{0}0.3^{0} (1-0.3)^{6-0}+ \binom{6}{1}0.3^{1} (1-0.3)^{6-1}+ \binom{6}{2}0.3^{2} (1-0.3)^{6-2}$

= $1 \times 1 \times 0.7^{6} +6 \times 0.3^{1} \times 0.7^{5} +15 \times 0.3^{2} \times 0.7^{4}$

= 0.11765 + 0.30253 + 0.32414 = 0.7443

Therefore, P(X < 3) = 0.7443.

5 0

3 years ago