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notsponge [240]
3 years ago
6

A value of a certain stock went up by 12% . Which functions can be used to find the current value of the stock that has an origi

nal value of x ? You may choose more than one correct answer. new value = original value + .12 f(x) = 1.12x f(x) = x + .12 new value = original value + .12(original value)
Mathematics
1 answer:
iris [78.8K]3 years ago
8 0
Let x = original value
12% = .12                    % means divide by 100
.12x = increase amount

f(x) = 1x + .12x
f(x) = 1.12x
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What is the slope and y-intercept of (5x-8y=3) Need help asap plz! thank you!
Ksenya-84 [330]
You can isolate y and you know that the slope is the number in front of x (the one that multiplies it) You know that the y-intercept is the value of y when x=0, therefore you can find y by replacing x with 0 in the equation 1-equation 5x-8y=3 5x=3+8y 5x-3=8y (5x-3)/8=y y=(5/8)x -3/8 (Slope = 5/8) 2- y-intecept (x=0) y=(5/8)(0)-3/8 y=-3/8
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3 years ago
A deck of cards contains 30 cards with labels 1, 2, . . . , 30. Suppose that somebody is randomly dealt a set of 7 cards of thes
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\displaystyle |\Omega|=\binom{30}{7}=\dfrac{30!}{7!23!}=\dfrac{24\cdot25\cdot\ldots\cdot30}{2\cdot3\cdot\ldots\cdot 7}=2035800

a)

\displaystyle\\|A|=\binom{15}{3}\cdot \binom{15}{4}=\dfrac{15!}{3!12!}\cdot\dfrac{15!}{4!11!}=\dfrac{13\cdot14\cdot15}{2\cdot3}\cdot\dfrac{12\cdot13\cdot14\cdot15}{2\cdot3\cdot4}=13\cdot7\cdot5\cdot13\cdot7\cdot15=621075\\\\P(A)=\dfrac{621075}{2035800}=\dfrac{637}{2088}\approx30.5\%

b)

\displaystyle\\|A|=\binom{10}{7}\cdot 3^7=\dfrac{10!}{7!3!}\cdot2187=\dfrac{8\cdot9\cdot10}{2\cdot3}\cdot2187=4\cdot3\cdot10\cdot2187=262440\\\\P(A)=\dfrac{262440}{2035800}=\dfrac{243}{1885}\approx12.9\%

4 0
2 years ago
In the first game of the year, the modified basketball team made 62.5% of their foul shot free throws. Matthew made all 6 of his
Lesechka [4]
So 6 shots equals 25% of the teams total shots(x).
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37.5% of 24= 24(0.375)=9

Hope this helps



5 0
4 years ago
The mean water temperature downstream from a power plant coolingtower discharge pipe should be no more than 102oF. Pastexperienc
nata0808 [166]

Answer:

(a) Yes, there is evidence that the water temperature is acceptable at \alpha = 0.05 (b) 0.9987 (c) 6.647274e-06

Step-by-step explanation:

Let X be the random variable that represents the water temperature. The water temperature has been measured on n = 9 randomly chosen days (a small sample), the sample average temperature is \bar{x} = 100°F and \sigma = 2°F. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

H_{0}: \mu = 102 vs H_{1}: \mu > 102 (upper-tail alternative)

We will use the test statistic

Z = \frac{\bar{X}-102}{2/\sqrt{9}} and the observed value is

z_{0} = \frac{100-102}{2/\sqrt{9}} = -3.

(a) The rejection region is given by RR = {z | z > 1.6448} where 1.6448 is the 95th quantile of the standard normal distribution. Because the observed value -3 does not belong to RR, we fail to reject the null hypothesis. In other words,  there is evidence that the water temperature is acceptable at \alpha = 0.05.  

(b) The p-value for this test is given by P(Z > -3) = 0.9987

(c) P(Accepting H_{0} when \mu = 106) = P(The observed value is not in RR when \mu = 106) = P(\frac{\bar{X}-102}{2/\sqrt{9}} < 1.6448 when \mu = 106) = P(\bar{X} < 102 + (1.6448)(2/\sqrt{9}) when \mu = 106) = P(\bar{X} < 103.0965) when \mu = 106) = P((\bar{X}-106)/(2/\sqrt{9}) < (103.0965-106)/(2/\sqrt{9}))) = P(Z < -4.3552) = 6.647274e-06

7 0
3 years ago
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