Im having trouble figuring out the missing length and solving this, can I have some help?

5 is subtracted from three-fifths of x...please someone translate this expression for me.

Fudgin [204]

3/5x-5
I believe this is the expression that is described.

8 0

3 years ago

At a large university, 20% of students are enrolled in the nursing program. The dean of students selects a random sample of 20 s

grin007 [14]

Answer:

He can select a row of the random number table and read 20 single digits between 1 and 5. He will record the number of digits that are a 1.

Step-by-step explanation:

edge 2022

3 0

2 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

3 years ago

Students at an elementary school are given a questionnaire that they are required to return after their parents have completed i

Nostrana [21]

Yes. The 100 - 85 = 15, meaning 15% of parents said yes. 85 > 15

6 0

3 years ago

"The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and f

yuradex [85]

Answer:

a

$P(X = 4 ) = 0.1876$

b

$P(X \le 4) = 0.8358$

Step-by-step explanation:

From the question we are told that

The proportion that has outstanding balance is p = 0.20

The sample size is n = 15

Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card

$X \ \ ~ Bin (n , p )$

Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as

$P(X = 4 ) = ^nC_4 * p^4 * (1 - p)^{n-4}$

=> $P(X = 4 ) = ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}$

Here C stand for combination

=> $P(X = 4 ) = 0.1876$

Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as

$P(X \le 4) = [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]$

=> $P(X \le 4) = 0.8358$

4 0

3 years ago