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Gelneren [198K]
1 year ago
15

Sue played four games of golf for these games her modal score was 98 and her mean score was 100 her range of score was 10 what w

as her score for the four games
Mathematics
1 answer:
earnstyle [38]1 year ago
8 0

Answer : Remaining two observation becomes 97 and 107.

Explanation :

Since we have given that

Mean = 100

Modal value = 98

Range = 10

As we know that ,

Range = Highest-Lowest

Let highest observation be x

Let lowest observation be y

So equation becomes x-y=10 ----equation 1

So, observation becomes

x,98,98,y

Now, we use the formula of mean i.e.

Mean = \frac{\text{Sum of observation}}{\text{N.of observaton}}

So, mean =\frac{x=98=98+y}{4}=400\\\frac{196+x+y}{4}=100\\x+y=400-196\\x-y=204

So our 2nd equation becomes

x+y=204

On using elimination method of system of linear equation on these two equation we get,

x=97

and

x+y=204\\y=204-x\\y=204-97\\y=107

Hence , remaining two observation becomes 97 and 107.

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katrin [286]

Answer:

27 years

Step-by-step explanation:

Given that :

Adrian's age = 3

Father's age = 34

In how many years will Adrian be twice as young as his father?

Let the number of years = x

2(3 + x) = 34 + x

6 + 2x = 34 + x

2x - x = 34 - 6

x = 28

3 0
1 year ago
Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that for at least one of
sasho [114]

Answer:

a)0,45119

b)1

Step-by-step explanation:

For part A of the problem we must first find the probability that both people in the couple have the same birthday (April 30)

P=\frac{1}{365} *\frac{1}{365}=\frac{1}{133225} \\

Now the poisson approximation is used

λ=nP=80000*1/133225=0,6

Now, let X be the number of couples that birth April 30

P(X ≥ 1) =

1 − P(X = 0) =

1-\frac{(e^-0.6)*(-0,6)^{0} }{0!}

P(X ≥ 1) = 0,45119

B)  Now want to find the

probability that both partners celebrated their birthday on th, assuming that the year is 52 weeks and therefore 52 thursday

P=52*\frac{1}{365} *\frac{1}{365}=\frac{52}{133225} \\

Now the poisson approximation is used

λ=nP=80000*52/133225=31.225

Now, let X be the number of couples that birth same day

P(X ≥ 1) =

1 − P(X = 0) =

1-\frac{(e^-31.225)*(-31.225)^{0} }{0!}

P(X ≥ 1) = 1

6 0
1 year ago
2. In Game 1, Emerson struck out
Dafna1 [17]
You would add the 30 to the 40 And then add 45 to that and you will get 112
7 0
1 year ago
The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Nataliya [291]

Answer:

(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716

(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;

    Mean, \mu = 30.05 inches        and    Standard deviation, \sigma = 0.2 inches

Let X = A sheet selected at random from the population

Here, the standard normal formula is ;

                  Z = \frac{X - \mu}{\sigma} ~ N(0,1)

(a) <em>The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65) </em>

P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)

P(X < 30.65) = P(\frac{X - \mu}{\sigma} < \frac{30.65 - 30.05}{0.2} ) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425

                                                                                                = 0.9985

P(X <= 30.25) = P( \frac{X - \mu}{\sigma} <= \frac{30.25 - 30.05}{0.2} ) = P(Z <= 1) = 0.84134

Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .

(b)<em> Let Y = Standard Normal Variable is given by N(0,1) </em>

<em> Which means mean of Y = 0 and standard deviation of Y = 1</em>

So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)

 P(Y < 3.2) = P(\frac{Y - \mu}{\sigma} < \frac{3.2 - 0}{1} ) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688

                                                                                           = 0.99931

 P(Y <= 0.3) = P(\frac{Y - \mu}{\sigma} <= \frac{0.3 - 0}{1} ) = P(Z <= 0.3) = 0.61791

Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .

 

3 0
1 year ago
2 Points
hichkok12 [17]

Answer: 345.02

Step-by-step explanation:

8 0
1 year ago
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