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3 years ago

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John Maynord pays a $566.00 annual insurance premium. This represents 40% of the total premium. The rest of the premium is paid

by his employer. Calculate:
. the company's percentage
. the total premium
. the company's payment

2 answers:

Aleksandr-060686 [28]3 years ago

6 0

60%
$1415
$849
Hope this helps

baherus [9]3 years ago

6 0

Answer:

A. 60%

B. $1415

C. $849.

Step-by-step explanation:

We have been given that John Maynord pays a $566.00 annual insurance premium. This represents 40% of the total premium. The rest of the premium is paid by his employer.

A. Since John pays 40% of his premium and rest of the premium is paid by his employer. This means that his employer pays 60% (100-40) of the premium, therefore, the company's percentage will be 60%.

B. Let x be the total amount of the premium.

As 40% of the premium is $566, so we need to find x such that 40% of x is $566.

We can represent this information in an equation as:

$\frac{40}{100}*x=\$566$

$0.40*x=\$566$

$\frac{0.40*x}{0.40}=\frac{\$566}{0.40}$

$x=\$1415$

Therefore, the total amount of premium is $1415.

C. As company pays 60% of premium, so to find the company's payment we will find 60% of $1415.

$\text{The company's payment}=\frac{60}{100}*\$1415$

$\text{The company's payment}=0.60*\$1415$

$\text{The company's payment}=\$849$

Therefore, the company's payment is $849.

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3 years ago

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

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Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

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Generally the degree of freedom for this data set is mathematically represented as

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substituting values

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$\alpha = 0.10$

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Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

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Equation of the line:

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or
y =
- 3 x
2
– -9
When x=0, y = -9
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Hope this helps you

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