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Gemiola [76]
2 years ago
9

2% of time a baseball player gets a double. 15% of the time a basketball player strikes out. What is the probability that the ba

seball player will not get a double?
2%

1%

99%

98%
Mathematics
1 answer:
telo118 [61]2 years ago
6 0
98 percent is the correct answer
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Answer:

33 is the answer

Step-by-step explanation:

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D. 21 m

Step-by-step explanation:

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2 years ago
Given: AB || CD
jarptica [38.1K]
First, find slope of point A and B using slope formula:
<u>y2-y1</u> = <u>7-0 </u>= <u>-7</u>
x2-x1    3-8     5

Next, use the point-slope formula to find the equation (pick either point A or B to substitute into this equation; the answer will be the same either way):
y-y1=m(x-x1)
y-7=<u>-7</u>(x-3)   (I used point B here)
        5
y=<u>-7x</u> +<u>56</u> so the y-intercept is <u>56</u>. Hurray! Part 1 down!
<span>     5      5                                   5

Now to answer part 2. Since AB ll CD, they have the same slope: <u>-7</u>
                                                                                                         5
Therefore, you can use the handy point-slope equation to calculate the equation of line CD. (Remember you only need one of the points to use this equation if you already have the slope.) Since the only point given is D(5,5), we'll use that one:
</span>y-y1=m(x-x1)
y-5=<u>-7</u>(x-5)
<span>        5
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<span>     5</span>
3 0
2 years ago
Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species
kotegsom [21]

Answer:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

Step-by-step explanation:

The logistic equation is the following one:

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that P(0) = 80, K = 2000.

The number of fish tripled in the first year. This means that P(1) = 3P(0) = 3(80) = 240.

Using the equation for P(1), that is, P(t) when t = 1, we find the value of r.

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}

280*(2000 + 80(e^{r} - 1)) = 160000e^{r}

280*(2000 + 80e^{r} - 80) = 160000e^{r}

280*(1920 + 80e^{r}) = 160000e^{r}

537600 + 22400e^{r} = 160000e^{r}

137600e^{r} = 537600

e^{r} = \frac{537600}{137600}

e^{r} = 3.91

Applying ln to both sides.

\ln{e^{r}} = \ln{3.91}

r = 1.36

This means that the expression for the size of the population after t years is:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

4 0
3 years ago
Solve: -8(x - 3y - 9)<br> Your answer
Verizon [17]
-8x + 24y + 72

————————-
6 0
2 years ago
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