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2 years ago

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if i have a 65.05% (D) in a class and i get a 100% on an assignment that worth 10% of my grade, what would my grade be

1 answer:

lord [1]2 years ago

4 0

I’m pretty sure you would have a 72% which would be a low C but maybe still passing. My schools passing grade is 70%. You would have to see what a passing grade is in your school.

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Maria rode her bike 8 miles south, then 3 miles east and then 2 miles

madam [21]

Answer:

6 miles north and 3 miles west

Step-by-step explanation:

If you picture these miles on a graph, you start at (0,0). Then you go 8 miles south (0, -8). Then 3 miles east (3, -8). Finally, 2 miles north (3, -6). You are 3 miles east and 6 miles south, so you go the opposite directions.

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3 years ago

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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3 years ago

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Arlecino [84]

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3 years ago

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Elias and Mark together have $756.80. Elias has $489.50. Write and solve an addition equation to determine how much money Mark (

Degger [83]

Answer:

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the last option is the correct option

Step-by-step explanation:

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