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bixtya [17]
2 years ago
14

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

Mathematics
1 answer:
belka [17]2 years ago
8 0
I think the answer is 17
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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.
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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
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