All categories

4 years ago

3 (x-5)?????????????

Mathematics

2 answers:

Yakvenalex [24]4 years ago

6 0

3(x-5), using the distributive property, we get 3x - 15

Sergeu [11.5K]4 years ago

4 0

Use the distributive property.

3*x - 3*5

3x - 15

Answer: 3x - 15

You might be interested in

The length of a rectangle is 3 1/6 cm longer than the width. The perimeter of the rectangle is 15 1/3 cm. What are the width and

eimsori [14]

Answer:

1212

Step-by-step explanation:

121212212

8 0

3 years ago

Which method can't you use to solve x²+7x=0.

scoray [572]

Answer:

Factoring

Step-by-step explanation:

x^2 + 7x = x(x + 7) = 0

x = 0 or x = -7

6 0

3 years ago

X^2+18x+72 factoring

Andre45 [30]

Answer:

92kyroydyodpgxtzoggz

Step-by-step explanation:

2+92r

3 0

3 years ago

How do I verify #15 using fundamental trig identities?

expeople1 [14]

We start with the more complicated side which is the left side, and show that, on using some trigonometric identities, we will get the term on the right side .

$\frac{sin \theta + tan \theta}{1+cos \theta}$

Using Quotient identity for tangent function, we will get

$\frac{sin \theta+ \frac{sin \theta}{cos \theta}}{1+cos \theta}$

$\frac{sin \theta cos \theta + sin \theta}{cos \theta(1+cos \theta)}$

Taking out sine function from the numerator

$=\frac{sin \theta(1+cos \theta)}{cos \theta(1+cos \theta)}$

Cancelling the common term of numerator and denominator

$=\frac{sin \theta}{cos \theta} = tan \theta$

7 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