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2 years ago

How do I solve the equation; 6(x^2-2)=12

Mathematics

1 answer:

anyanavicka [17]2 years ago

4 0

Ok so first we simplify 6(x^2-2)=12 6*x^2 = 6^2 and 6*-2 = -12 = 6x^2 -12 = 12 do you understand?

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Find the measure of angle D and E Angle A = 90 degrees Angle B = 90 degrees Angle C= Angle D Angle E = 90 degrees

Kobotan [32]

Answer:

Step-by-step explanation:

all angles have to add up to a total of 360, so the other 3 angles equal 270 and there's two more angles that need to split 90° equally

5 0

3 years ago

ruth had a ribbon that was 9 yards long. she cut the ribbon into pieces that were 1/6 yard long. how many pieces of ribbon does

Ainat [17]

Answer:

Step-by-step explanation:

6 x 9 = 54

she can there are

make 6 9 yards

ribbons

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a yard

4 0

3 years ago

What is the product of 3x(x2 + 4)? o x2 + 3x + 4 3r3+ 4 O 3x + 12x o 3x2 + 12x

Ipatiy [6.2K]

Answer:

Step-by-step explanation:

3x(x^2+4)

=3x^2+12x

8 0

3 years ago

) 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

I need help with a math question but apparently everyone is dead today..

Gwar [14]

The answer is D. I’m assuming this is surface area so here’s my explanation:) ok so the area of the triangle is 6 (4x3 divided by 2) and when you add the other triangle that’s 12. Ok so we have 12 so far. Then the area of the bottom rectangle is 24 (3x8). Then the area of the side rectangle is 32 (4x8). THEN the top rectangle’s area is 40 (8x5). When you add the numbers up (12+24+32+40) you get 108! And of course dont forget the cm2. ( centimeters squared ). :) I hope this helps

8 0

2 years ago