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1 year ago

Write a paragraph of a real life situation you can use the graph as and explain how u would fit it onto the graph.

Mathematics

1 answer:

ivann1987 [24]1 year ago

3 0

We can use this graph in the situation of an animal population of an animal species that grows exponentially until the death of Jesus, and from that point began to decrease exponentially.

It fit in the graph of the following way:

The y-axis represents the number of animals, and the x axis represents the passing of time, 0 becomes the moment in which the calendar stops being taken as before Christ and starts to be taken as after Christ.

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Simplify the expression-14-(-8)

Y_Kistochka [10]

Answer:

the answer is -6

Step-by-step explanation:

-14-(-8)=-6.

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3 years ago

Write in point-slope form an equation of the line passing through (-4, 10) and (-6, 15). Show your work.

Neporo4naja [7]

This is how it is done. Let me know if you have questions...

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3 years ago

How much interest does $400 make in 5 years at 7% per annum?

ZanzabumX [31]

Answer:

$112

Step-by-step explanation:

7% for each year = $28

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3 years ago

A manufacturer of golf clubs makes a profit of $60 per set on a model A set and $45 per set on a model B set. Daily production o

KengaRu [80]

Answer:

If the question does not limit man to produce only one model of club, then the maximized profit of every condition produced under 50 sets daily will be all model A exclusive. such as $60 x 50 (model A) , even just produce 49 set that day, the maximal profit is still $60 x 49.

Step-by-step explanation:

Re-consider the logic of the question ....

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3 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = population proportion.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, the test statistics = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that this is an unusually high number of faulty modems.

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3 years ago