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3 years ago

16 + 4x + 2x-3x + x = 0

2 answers:

4 0

16 + 4x + 2x - 3x + x = 0

First, simplify 16 + 4x + 2x - 3x + x to get 16 + 4x. / Your problem should look like: 16 + 4x = 0
Second, subtract 16 from both sides. / Your problem should look like: 4x = -16
Third, divide both sides by 4. / Your problem should look like: x = $-\frac{16}{4}$
Fourth, simplify $\frac{16}{4}$ to get 4. / Your problem should look like: x = -4

Answer: x = -4

nekit [7.7K]3 years ago

4 0

16+4x+2x−3x+x=0

Simplify:

16+4x+2x−3x+x=0

16+4x+2x+−3x+x=0

(4x+2x+−3x+x)+(16)=0(Combine Like Terms)

4x+16=0

Subtract 16 from both sides.

4x+16−16=0−16

4x=−16

Divide both sides by 4.

4x/4=−16/4

x=−4

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) The National Highway Traffic Safety Administration collects data on seat-belt use and publishes results in the document Occupa

asambeis [7]

Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

Let $p_1$ = population proportion of drivers 16-24 years old who buckle up .

$p_2$ = population proportion of drivers 25-69 years old who buckle up .

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no significant difference in seat belt use}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a difference in seat belt use}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers 16-24 years old who buckle up = 79%

$\hat p_2$ = sample proportion of drivers 25-69 years old who buckle up = $\frac{924}{1100}$ = 84%

$n_1$ = sample of 16-24 years old drivers = 1000

$n_2$ = sample of 25-69 years old drivers = 1100

So, test statistics = $\frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }$

= -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics doesn't lie within the range of critical values of z, 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 there is a difference in seat belt use.

4 0

3 years ago

9. A ladder of length 23 feet leans against the side of a building. The angle of elevation of the ladder is 76. Find the distanc

irakobra [83]

<h3>Answer:</h3>

B) 22.32 feet

$\quad\rule{300pt}{1pt}$

<h3>Solution:</h3>

we are given that , a ladder is placed against a side of building , which forms a right angled triangle . We wre given one side of a right angled triangle ( hypotenuse ) as 23 feet and the angle of elevation as 76 ° . We can find the Perpendicular distance from the top of the ladder go to the ground by using the trigonometric identity:

$\qquad\quad\bull ~{\boxed{\bf{ Sin\theta =\dfrac{Perpendicular}{Hypotenuse}}} }~\bull$