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2 years ago

Which exponential function is graphed here?

Mathematics

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

y = 5^x

Step-by-step explanation:

y= b*(a)^x + c

c could = 1 but then you would not have an exponential function. c = 0 because the graph follows the x axis up until x = -2. Suppose c = 1. The the graph would follow y = 1 up until x = - 2

When x = 0, y = 1 which means that b. If b is anything but 0 or 1 then the y intercept would be stretched to a different place. If be = 0 then y would = 0.

So the graph is of the form y = a^x

Now when x = 0 the graph, the y intercept is y = a^0 or y = 1 So the y intercept is (0,1)

Now the next point is thing to solve for is a.

When x = 1, y = 5 (read the graph)

y = a^x

5 = a^1

5 = a because a^1 is a.

Answer

y = 5^x.

5 = a^1

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2 years ago

A basket of the fruit contains 12 apples, 8 bananas, and 6 oranges. What is the ratio of bananas to apples? (Please simplify)

mrs_skeptik [129]

Answer:

2 : 3

Step-by-step explanation:

bananas to apples = 8 : 12

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2 years ago

What is 12 more than 8.2 times a number n

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3 years ago

2(y - 6) = 3ty - 4) - y

Serggg [28]

Answer:

80000

Step-by-step explanation:

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2 years ago

EU (European Union) countries report that 46% of their labor force is female. The United Nations wants to determine if the perce

miss Akunina [59]

Answer:

b. We are 95% confident that between 35.5% and 44.3% of the persons in the U.S. labor force is female.

Step-by-step explanation:

1) Data given and notation

n=500 represent the random sample taken

X represent the number of females in the U.S. labor force

$\hat p=0.46$ estimated proportion of females in the U.S. labor force

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic

p= population proportion of females in the U.S. labor force

2) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.46 - 1.96 \sqrt{\frac{0.46(1-0.46)}{500}}=0.416$

$0.46 + 2.58 \sqrt{\frac{0.46(1-0.46)}{500}}=0.504$

On this case the calculated interval is not the given , but let's assume that the confidence interval is given by the statment: "United States Department of Labor find that 95% confidence interval for the proportion of females in the U.S. labor force is .357 to .443."

3) Correct interpretation

a. The margin of error for the true percentage of females in the U.S. labor force is between 35.7% and 44.3%.

False the confidence interval not conclude about the margin of error conclude about the true proportion.

b. We are 95% confident that between 35.5% and 44.3% of the persons in the U.S. labor force is female.

True, the statement report the confidence level and the limits for the margin of error.

c. The percentage of females in the U.S. labro force is between 35.7% and 44.3%.

False, The statement is not correct because not reports the confidence level.

d. All sample of size 500 will yield a percentage of females in the U.S. labor force that falls within 35.7% and 44.3%

False the confidence interval is for the population proportion not just for the samples of size 500

e. None of these.

False, we have an option that is true.

8 0

2 years ago