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3 years ago

How do I find the zeros using factoring

Mathematics

1 answer:

Lynna [10]3 years ago

6 0

To much of an open ended question

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Whats the variable of 1

Fudgin [204]

Answer:

There are no variables, there is only a 1

Step-by-step explanation:

7 0

3 years ago

True or false: equilateral triangles can be classified as acute right or obtuse

Ne4ueva [31]

False, since all sides are equal all angles must be equal and when you have an obtuse or right triangle only on angle is obtuse or right. since one is different it cannot be equal on all sides making this statement false

8 0

3 years ago

A news report suggested that an adult should drink a minimum of 4 pints of water per day.

spin [16.1K]

9514 1404 393

Answer:

448

Step-by-step explanation:

(4 pt/da)×(7 da/wk)×(16 oz/pt) = 448 oz/wk

That would be 448 fluid ounces per week.

3 0

3 years ago

Find FG FGH= 49 GH=31

Luba_88 [7]

Answer:

Step-by-step explanation:

From the given diagram;

FGH = 49

GH = 31

Using the expression to get FG;

FG + GH = FGH

FG + 31 = 49

FG = 49-31

FG = 18

Hence the measure of FG is 18

4 0

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.

6 0

3 years ago