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gulaghasi [49]
3 years ago
14

Abbie, Brad, Cindy, and Dan went bowling one night. For fun, they decided to team up the girl's against the boys, with the highe

st total score for the each team winning. Abbie scored 16 more than Brad. Cindy's score was 43 less than Dan's. Brad outscored Dan by 23. It the total scored by all four kids was 599, what was each person's score and what was the final score for the match?
Mathematics
1 answer:
Lemur [1.5K]3 years ago
3 0
I love this app but I love you so I love you too baby I love it soo many of my followers are you utieiu
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What is the exact volume of the cylinder?
rewona [7]
128pi in^3

because:
Volume = Area of the base * height
Volume= pi*r^2*height
= pi4^2 * 8
= 128pi
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2 years ago
Solve the simultaneous equations<br> 7x – 3y = 24<br> 2x + 3y = 3<br> x=?<br> y =?
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Answer:

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Step-by-step explanation:

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3 years ago
Two candidates ran for class president. The candidate that won received 60%of the 200 total votes. How many votes did the winnin
Zanzabum

Answer: 120

Step-by-step explanation: You multiply 200 and .6 and you should get 120.

8 0
3 years ago
Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. use the 68-95-99.7 rule to
Novosadov [1.4K]

Answer:

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

Step-by-step explanation:

We define the random variable representing the prices of a certain model as X and the distirbution for this random variable is given by:

X \sim N(\mu = 150000, \sigma =2300

The empirical rule states that within one deviation from the mean we have 68% of the data, within 2 deviations from the mean we have 95% and within 3 deviations 99.7 % of the data.

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

7 0
3 years ago
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