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3 years ago

What is the value of the 100th term in this arithmetic sequence? 5, 7, 9, 11, ...

Mathematics

1 answer:

OLEGan [10]3 years ago

3 0

Answer:

the 100th term would be 193

Step-by-step explanation:

5-(n-1) times 2

5-99 times 2

5-198

193

I hope this helps

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A jar contains 21 yellow marbles and 22 orange marbles. A marble is drawn at random. P(orange)

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1/43 is the probability of drawing an orange marble.

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Ok, we know that we can write a horizontal translation as:

y' = f(x - A)

where if A is positive, this moves the graph of f(x) A units to the right.

Why is this?

Ok, let's compare:

y = f(x)

and

y' = f(x - A)

in y, when x = 0 we have f(0).

While to have this same point in y', we need to evaluate in x = A.

f(A - A) = f(0).

Then the value f(0) is now at x = A, this means that the point moved A units to the right.

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3 years ago

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The ratio of red pens to blue pens in a package is 2:5. If there are 35 blue pens in the package, how many red pens are in t

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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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4 years ago