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3 years ago

Make the following changes to the number 5.376: Make the 3 worth 10 times as much. Makethe5worth_1 asmuch.

Mathematics

1 answer:

cluponka [151]3 years ago

3 0

Answer:

Step-by-step explanation:

your mom

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In this diagram which equation could you prove to be true in order to conclude that the lines are parallel

KiRa [710]

I would say that the correct answer is D. because y is always the numerator while x is the denominator for the equation y2 - y1/x2 - x1 which means if there is two y's or two x's on the same line you subtract the second one from the first one. If there is only one y and/or x and the other is 0 on the same line, it stays at y or x without subtracting y2 - y1 or x2 - x1.

Since b is on the y coordinate and -a is on the x coordinate, you would make it b/a while -a is gonna be a positive since the lines are going up and to the right. Now, since c is the y coordinate and d is the x coordinate, make C the numerator and d the denominator since y is always the numerator and x is the denominator for these parallel line figures on the graph and the equation will be equaled to the fraction to the other fraction for parallel lines.

So, your answer would be D. b/a = c/d

Hope this helps, and is correct!

~ ShadowXReaper069

3 0

3 years ago

Read 2 more answers

Please solve completely I really need this answered

Harrizon [31]

Answer:

Ted is correct. Maggie made mistakes while trying to isolate x for both equations.

Step-by-step explanation:

For 3x-2=0, in order to move -2 to the left side, Maggie had to add 2 on both sides because -2+2=0. The same problem is seen for x+5=0. Maggie had to subtract 5 on both sides because +5-5=0

3 0

3 years ago

Please help me on #37. I know for A you would use the quadratic equation but I got the wrong answer.

ozzi

Δ = (8i)^2 - 4*(-25) => Δ = -36 +100 => Δ = 64 => x1 =(-8i + 8)/2 => x1 = -4i +4;
x2 = (-8i - 8)/2 => x2 = -4i-4; in this case, your solutions are complex conjugates.

6 0

3 years ago

1. Find the distance from point P to QS

12345 [234]

Answer:

5.7 units

Step-by-step explanation:

The distance from point P to QS is the distance from point P (1, 1) to the point of interception R(-3, 5).

Use distance formula to calculate distance between P and R:

$PR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$P (1, 1) = (x_1, y_1)$

$R(-3, 5) = (x_2, y_2)$

Plug in the values into the formula.

$PR = \sqrt{(-3 - 1)^2 + (5 - 1)^2}$

$PR = \sqrt{(-4)^2 + (4)^2}$

$PR = \sqrt{16 + 16}$

$PR = \sqrt{32}$

$PR = 5.7 units$ (to nearest tenth)

4 0

3 years ago

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple

polet [3.4K]

Answer:

95% confidence interval for the difference in the proportion is [-0.017 , 0.697].

Step-by-step explanation:

We are given that a simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 8 were "totaled," meaning that the cost of repairs is greater than the value of the car.

Another sample of 15 large cars were subjected to the same test, and 5 of them were totaled.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

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

where, $\hat p_1$ = sample proportion of small cars that were totaled = $\frac{8}{12}$ = 0.67

$\hat p_2$ = sample proportion of large cars that were totaled = $\frac{5}{15}$ = 0.33

$n_1$ = sample of small cars = 12

$n_2$ = sample of large cars = 15

$p_1$ = population proportion of small cars that are totaled

$p_2$ = population proportion of large cars that were totaled

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between population population, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < $p_1-p_2$ < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

95% confidence interval for $p_1-p_2$ = [ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ]

= [ $(0.67-0.33)-1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ , $(0.67-0.33)+1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ ]

= [-0.017 , 0.697]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].

6 0

4 years ago