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Alchen [17]
3 years ago
8

The graph of y = -3x + 2 intersects the graph of an exponential function at (-1, 5). Which of the following could be the exponen

tial function?
a) y = 10(2)
^{x}
b) y = 10(0.5)^{x}
c) y = 5^{x}
d) y = 5(2)^{x}
Mathematics
2 answers:
Andre45 [30]3 years ago
8 0
Of the function choices offered, only the first includes (-1, 5) in its graph.
  a) y = 10(2)^x
miskamm [114]3 years ago
7 0

Answer:

a) y = 10(2)^x

Step-by-step explanation:

In order to solve this you just need to evaluate the options given with the values of the point that is given in order to see if that point is a possibility in that function, remember to use the value of x to avaluate and if the result is the same as the points value for y then that poin is part of the graph.

a) y = 10(2)^x

y=10(2)^x\\y=10(2)^-1\\y=10*\frac{1}{2} \\y=5

Remember that when elevating to negative numbers the ecponent becomes a numerator, so the only option that is correct would be the first one a) y = 10(2)^x

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➷ Just substitute 3 in:

2(3)^2 = 18

It would be 18

<h3><u>✽</u></h3>

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4 years ago
Read 2 more answers
Test the claim that the proportion of people who own cats is significantly different than 80% at the 0.2 significance level. The
sattari [20]

Answer:

C. H0 : p = 0.8 H 1 : p ≠ 0.8

The test is:_____.

c. two-tailed

The test statistic is:______p ± z (base alpha by 2) \sqrt{\frac{pq}{n} }

The p-value is:_____. 0.09887

Based on this we:_____.

B. Reject the null hypothesis.

Step-by-step explanation:

We formulate null and alternative hypotheses as  proportion of people who own cats is significantly different than 80%.

H0 : p = 0.8 H 1 : p ≠ 0.8

The alternative hypothesis H1 is that the 80% of the  proportion is different and null hypothesis is , it is same.

For a two tailed test for significance level = 0.2 we have critical value  ± 1.28.

We have alpha equal to 0.2  for a two tailed test . We divided alpha with 2 to get the answer for a two tailed test. When divided by two it gives 0.1 and the corresponding value is ± 1.28

The test statistic is

p ± z (base alpha by 2) \sqrt{\frac{pq}{n} }

Where p = 0.8 , q = 1-p= 1-0.8= 0.2

n= 200

Putting the values

0.8 ± 1.28 \sqrt{\frac{0.8*0.2}{200} }

0.8 ± 0.03620

0.8362, 0.7638

As the calculated value of z lies within the critical region  we reject the null hypothesis.

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3 years ago
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6.665 grams of the 13 grams remain after 8 hours.

<h3>How much of a 13 gram sample of iron-52 would remain after 8 hours?</h3>

The decay equation for the 13 grams of iron-52 is:

N(t) = 13g*e^{(-ln(2)/8.3h)*t}}

Where N is the amount of iron-52, and t is the time in years.

Where we used the fact that the half-life is exactly 8.3 hours.

Now, the amount that is left is given by N(8h), so we just need to replace the variable t by by 8 hours, so we get:

N(8h) = 13g*e^{(-ln(2)/8.3h)*8h}} = 6.665g

So 6.665 grams of the 13 grams remain after 8 hours.

If you want to learn more about half-life:

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Sinθ/cosθtanθ=1<br>how is this identity true?​
Yakvenalex [24]

Answer:

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Which of the following would be an acceptable first step in simplifying the expression
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Answer: A

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