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bonufazy [111]
2 years ago
14

Assignment Mode : Open

Mathematics
2 answers:
krok68 [10]2 years ago
8 0
The answer is 300 divided by 20 which you would get 15.
hodyreva [135]2 years ago
8 0
15 mpg is the answer
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To calculate the number of montly payments required to pay off a loan or meet an investment goal, use the _______ function.
soldier1979 [14.2K]
NPER function

Hope this helps... mark as Brainliest plz
7 0
3 years ago
If the original square had a side length of
irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas  in the attached figure N 3, and the trinomial is x^{2} +11x+28

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

Length=(x+4)\ in

width=(x+7)\ in

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas  in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

A1=(x)(x)=x^{2}\ in^2

A2=(4)(x)=4x\ in^2

A3=(x)(7)=7x\ in^2

A4=(4)(7)=28\ in^2

The total area of the second rectangle is the sum of the four areas

A=A1+A2+A3+A4

State the product of (x+4) and (x+7) as a trinomial

(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28

Part c) If the original square had a side length of  x = 2 inches, then what is the area of the  second rectangle?

we know that

The area of the second rectangle is equal to

A=A1+A2+A3+A4

For x=2 in

substitute the value of x in the area of each portion

A1=(2)(2)=4\ in^2

A2=(4)(2)=8\ in^2

A3=(2)(7)=14\ in^2

A4=(4)(7)=28\ in^2

A=4+8+14+28

A=54\ in^2

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

A(x)=x^{2} +11x+28

For x=2 in

substitute and solve for A(x)

A(2)=2^{2} +11(2)+28

A(2)=4 +22+28

A(2)=54\ in^2 ----> verified

therefore

The trinomial represent the total area of the second rectangle

7 0
3 years ago
3 The table shows the prices of some breakfast items at a restaurant. Sara ordered 2 eggs, a
AleksandrR [38]
It is B just use a calculator hope it helped
6 0
2 years ago
What is the negation of this statement x+y 10?
andre [41]
I'm not sure what your statement is.
Is it x+y = 10?

Because if it is, the negation of this statement would simply be x+y<span>≠10, which stands for not equal. </span>
7 0
2 years ago
A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall
Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : \mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7

Standard deviation : \sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}

=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}

=\dfrac{30.265}{10}=3.0265

The confidence interval for the population mean (for sample size <30) is given by :-

\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given significance level : \alpha=1-0.95=0.05

Critical value : t_{n-1,\alpha/2}=t_{9,0.025}=2.262

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0
3 years ago
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