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beks73 [17]
3 years ago
9

Gabriela walked to a toy store around noon and , after browsing for 2222 minutes, decided to buy a race car for $9.22. Gabriela

handed the salesperson $9.22 for her purchase.
How much change did Gabriela receive?
Mathematics
1 answer:
Yanka [14]3 years ago
5 0
No change. She gave the salesperson exactly the right amount to buy the racecar.
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The length of the base of an isosceles triangle is x. The length of a leg is 4x - 2. The perimeter of the triangle is 95. Find x
adell [148]

Answer:

x = 11

Step-by-step explanation:

sides of triangle measure: 'x', '4x-2' and '4x-2' since isosceles means both legs are equal

perimeter is the sum of all sides, so set up equation:

95 = 4x - 2 + 4x - 2 + x

95 = 8x - 4 + x

95 = 9x - 4

99 = 9x

x = 11

5 0
3 years ago
Wouldn't (D) be the right answer?? I divided 23% by 5,260, but I'm not 100% sure it's correct.
atroni [7]
It would be c. Ian would predict that 23% of the 5,260 would list a sport. Usually when you phrase things like x OF y you multiply. So 23%(or .23) x 5,260 = 1209.8 and we can round to a whole person, giving us 1,210 people
3 0
3 years ago
Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
A production process is known to produce a particular item in such a way that 5 percent of these are defective. If two items are
IRINA_888 [86]

Answer:

0.0025 = 0.25% probability that both are defective

Step-by-step explanation:

For each item, there are only two possible outcomes. Either they are defective, or they are not. Items are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

5 percent of these are defective.

This means that p = 0.05

If two items are randomly selected as they come off the production line, what is the probability that both are defective

This is P(X = 2) when n = 2. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025

0.0025 = 0.25% probability that both are defective

4 0
2 years ago
Find the midpoint of the segment ending in the points 0,9 and 5,1
Finger [1]
U se the midpoint formula to find the midpoint of the line segment.
(5/2,5) or (5,5/2) it’s one of them .gotta be I’m pretty sure it’s (5/2,5)
4 0
3 years ago
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