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3 years ago

6

Henry knows that the area of a rectangle is 30 square inches. The perimeter is 22 inches. If the length is 1 inch longer than th

e width, what are the length and width of Henry's rectangle? Explain how you know.

1 answer:

dsp733 years ago

5 0

The length is 6 inches and the width is 5 inches

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How do the graphs of the functions f(x) = (Three-halves)x and g(x) = (Two-thirds)x compare?

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Answer:

The answer to this question can be defined as follows:

Step-by-step explanation:

Given:

$\bold{f(x)= (\frac{3}{2})^x}\\\\\bold{g(x)= (\frac{2}{3})^x}\\$

Following are the graph attachment to this question:

The second function, that is $g(x)= (\frac{2}{3})^x$ is not even a function.

Remember that g(x) function is the inverted f(x) function. And when you see this pattern, a reflection on the Y-axis expects you.

Reflection in the axis.

In x-axis:

Increase the function performance by -1 to represent an exponential curve around the x-axis.

In y-axis:

Increase the input of the function by -1 to represent the exponential function around the y-axis.

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Answer: 2

Step-by-step explanation: 1+1=2

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Help me answer 2/9 divided by 1/6

Zigmanuir [339]

1.33333

Step-by-step explanation:

Don't worry it checks out.

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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

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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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