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grandymaker [24]
3 years ago
9

Base Angle of an isosceles trapeziod are A.supplementary B.complementary C.congruent D.adjacent?

Mathematics
2 answers:
Art [367]3 years ago
8 0

Answer:

Congruent

Step-by-step explanation:

by definition, the base angles of an isosceles triangle are the same value.

i.e they are congruent.

antiseptic1488 [7]3 years ago
8 0
C. Your answer will be Congruent!
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The marginal price per pound (in dollars) at which a coffee store is willing to supply x pounds of Jamaican Blue Mountain coffee
il63 [147K]

Answer:

Thus, the coffee shop is willing to supply 6 pounds per week at a price of $4 per pound.

Step-by-step explanation:

We are given the following information in the question:

The marginal price per pound (in dollars) is given by:

p'(x) = \displaystyle\frac{208}{(x+7)^2}

where x is the supply in pounds.

P(x) = \displaystyle\int p'(x)~dx =\displaystyle\int\displaystyle\frac{208}{(x+7)^2}~dx\\\\P(x) = \frac{-208}{(x+7)} + c\\\\\text{where c is the constant of integration.}

The coffee shop is willing to supply 9 pounds per week at a price of $7 per pound.

Thus, we are given that

P(9) = 7

Putting the values, we get,

P(x) = \displaystyle\frac{-208}{(x+7)} + c\\\\P(9) = 7\\\\\displaystyle\frac{-208}{(9+7)} + c = 7\\\\c = 7 + \frac{208}{16} = 20

P(x) = \displaystyle\frac{-208}{(x+7)} + 20

Now, we have to find how many pounds it would be willing to supply at a price of $4 per pound.

P(x) = 4

P(x) = \displaystyle\frac{-208}{(x+7)} + 20 = 4\\\\\frac{-208}{x+7} = -16\\\\x + 7 = 13\\x = 6

Thus, the coffee shop is willing to supply 6 pounds per week at a price of $4 per pound.

3 0
3 years ago
IF NR = 8 ft, what is the length of arc NMP. Round to the nearest tenth.
Pavlova-9 [17]

Check the picture below.

since chords NQ and MP cross the center of the circle at R, that means that those two chords are diametrical chords and the angles made by both are vertical angles and thus twin angles, namely both are 18° as you see in the picture, so the angle NMP in magenta is really 162° + 18° + 18° = 198°, and we know the radius NR is 8.

\textit{arc's length}\\\\ s=\cfrac{r\pi \theta }{180}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=8\\ \theta =198 \end{cases}\implies s=\cfrac{(8)\pi (198)}{180}\implies s\approx 27.6

4 0
2 years ago
A child has 4 wooden blocks. How many different ways can she stack 3 of them into a tower?
faltersainse [42]

Answer:

Step-by-step explanation:

To find the number of different ways she can stack 3 of them in a tower, we need to use the formula:

_{n}P_{k}=\dfrac{n!}{(n-k)!}

n = 4

k = 3

_{n}P_{k}=\dfrac{4!}{(4-3)!}

_{n}P_{k}=\dfrac{4!}{1!}

_{n}P_{k}=\dfrac{4*3*2*1!}{(1)!}

_{n}P_{k}=\dfrac{4*3*2*1!}{1}

_{n}P_{k}=\dfrac{24}{1}

7 0
2 years ago
The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.35 en
trasher [3.6K]

Answer:

a) 0.3571

b) The p-value is 0.362007.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 1.35

Sample mean, \bar{x} = 1.4

Sample size, n = 26

Alpha, α = 0.01

Sample standard deviation, s = 0.7

First, we design the null and the alternate hypothesis

H_{0}: \mu = 1.35\text{ entrees per order}\\H_A: \mu > 1.35\text{ entrees per order}

We use One-tailed t test to perform this hypothesis.

a) Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{1.4 - 1.35}{\frac{0.7}{\sqrt{25}} } = 0.3571

b) The p-value at t-statistic 0.3571 and degree of freedom 25 is 0.362007.

4 0
3 years ago
Based on the survey, what is the probability that a person chosen at random is a diabetic patient or an eye patient
KonstantinChe [14]

First, you must find the total number of people. Add together 32, 54, 78, 112, and 96 to get 372. Next, add the number of diabetic patients with the number of patients with eye problems (54 + 112 = 166).

Your fraction is now 116/372. Simplify this to get your answer, 29/93 or 0.31.

6 0
2 years ago
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