The cost of a jacket increased from $85.00 to $99.45. What is the percentage increase of the cost of the jacket?

If the boy is 5’6 tall and his shadow is 4 ft and the shadow of the pole 18 ft determine the height of the pole to the nearest 1

Dominik [7]

The answer is 24.7ft

7 0

3 years ago

Can someone please help me with this

k0ka [10]

Answer: $3x^6+x^5-x^4-6x^3+x^2+3x-1$

Step-by-step explanation:

(3x^2+x-1)(x^4-2x+1)

=3x^6-6x^3+3x^2+x^5-2x^2+x-x^4+2x-1

=3x^6+x^5-x^4-6x^3+x^2+3x-1

Hope this helps!! :)

Please let me know if you have any question

5 0

3 years ago

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Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago

Find the value of x.

elena-s [515]

Hi there!

Using the formula - n-2*180 -, we know that the interior angles of a pentagon add up to 540. We can use the values we're given and add them together and set them all equal to 540, then solve.

WORK:
540 = 204 + 12x
336 = 12x
x = 28

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

8 0

3 years ago

Terry has just purchased a new car, which had a list price of $16,825. She had to pay 7.19% sales tax, a $1,128 vehicle registra

Anarel [89]

Total cost of purchasing the car = $16,825 + 0.0719 x $16,825 + $1,128 + $190 = $19,352.72
Value of her previous car = 0.9 x $16,825 = $15,142.50

Amount being owed by Terry = $19,352.72 - $15,142.50 = $4,210.22

The present value of annuity is given by PV = P(1 - (1 + r/t)^-nt) / (r/t)
where P is the monthly payment, r is the rate = 10.59% = 0.1059, t is the number of periods in a year = 12, n is the number of years.

4,210.22 = P(1 - (1 + 0.1059/12)^-(5 x 12)) / (0.1059/12)
0.1059(4,210.22) = 12P(1 - (1.008825)^-60)
P = 445.862298 / 4.916774288 = 90.68

Amount of interest paid = 60(90.68) - 4,210.22 = 5440.80 - 4210.22 = $1230.58

7 0

3 years ago

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