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Morgarella [4.7K]
2 years ago
12

Keith played the first 22 minutes of soccer game lohan then replaced him for the rest of half logan started the second half and

was replaced by wilson with 18 minutes left in the game if each half is 40 minutes ling how long did logan play during the second half
Mathematics
1 answer:
insens350 [35]2 years ago
6 0

Answer:

Logan played for 22 minutes during the second half.

Step-by-step explanation:

Since Keith played the first 22 minutes of a soccer game and Logan then replaced him for the rest of the half, and Logan started the second half and was replaced by Wilson with 18 minutes left in the game, if each half is 40 minutes long To determine how long did Logan play during the second half, the following calculation must be performed:

Second Half Total - Time Played by Wilson = Time Played by Logan

40 - 18 = X

22 = X

Therefore, Logan played for 22 minutes during the second half.

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3 years ago
How do I solve this?
meriva
Hello,

If 3x-1>=0 then

x>=1/3
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7 0
3 years ago
A group of students weighs 500 US pennies. They find that the pennies have normally distributed weights with a mean of 3.1g and
inysia [295]
Question 1

probability between 2.8 and 3.3

The graph of the normal distribution is shown in the diagram below. We first need to standardise the value of X=2.8 and value X=3.3. Standardising X is just another word for finding z-score

z-score for X = 2.8
z= \frac{2.8-3.1}{0.41} =-0.73 (the negative answer shows the position of X = 2.8 on the left of mean which has z-score of 0)

z-score for X = 3.3 
z= \frac{3.3-3.1}{0.41}=0.49

The probability of the value between z=-0.73 and z=0.49 is given by 
P(Z<0.49) - P(Z<-0.73)

P(Z<0.49) = 0.9879
P(Z< -0.73) = 0.2327 (if you only have z-table that read to the left of positive value z, read the value of Z<0.73 then subtract answer from one)

A screenshot of z-table that allows reading of negative value is shown on the second diagram

P(Z<0.49) - P(Z<-0.73) = 0.9879 - 0.2327 = 0.7552 = 75.52%

Question 2
Probability between X=2.11 and X=3.5

z-score for X=2.11
z= \frac{2.11-3.1}{0.41}=-2.41

z-score for X=3.5
z= \frac{3.5-3.1}{0.41} =0.98

the probability of P(Z<-2.41) < z < P(Z<0.98) is given by
P(Z<0.98) - P(Z<-2.41) = 0.8365 - 0.0080 = 0.8285 = 82.85%

Question 3
Probability less than X=2.96

z-score of X=2.96
z= \frac{2.96-3.1}{0.41}=-0.34
P(Z<-0.34) = 0.3669 = 36.69%

Question 4
Probability more than X=3.4
z= \frac{3.4-3.1}{0.41}=0.73
P(Z>0.73) = 1 - P(Z<0.73) = 1-0.7673=0.2327 = 23.27%



7 0
3 years ago
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