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2 years ago

How much do the cars weigh? URGENT! I need this before tomorrow!

Mathematics

1 answer:

faltersainse [42]2 years ago

6 0

Answer:

Here is the short answer. The U.S. Environmental Protection Agency states that the average weight of a car in 2018 was 4,094 pounds.

Step-by-step explanation:

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4^12*6^15*7^21<br><br> HEELPOOOOPO

Varvara68 [4.7K]

9514 1404 393

Answer:

4^4·6^5·7^7 = 1,639,390,814,208

Step-by-step explanation:

Taking the cube root multiplies each exponent by 1/3.

((4^12)(6^15)(7^21))^(1/3) = (4^(12/3))·(6^(15/3))·(7^(21/3)) = (4^4)(6^5)(7^7)

6 0

2 years ago

AN AIRCRAFT FLYING WITH THE WIND FLIES TO

maxonik [38]

Answer:

643

Step-by-step explanation:

3 0

3 years ago

Determine the equation of the following polyromial that passes through point (0,6).

Ostrovityanka [42]

Answer:

Cubic polynomial has zeros at x=−1x=−1 and 22, is tangent to x−x−axis at x=−1x=−1, and passes through the point (0,−6)(0,−6).

So cubic polynomial has double zero at x=−1x=−1, and single zero at x=2x=2

f(x)=a(x+1)2(x−2)f(x)=a(x+1)2(x−2)

f(0)=−6f(0)=−6

a(1)(−2)=−6a(1)(−2)=−6

a=3a=3

f(x)=3(x+1)2(x−2)f(x)=3(x+1)2(x−2)

f(x)=3x3−9x−6

3 0

2 years ago

Which equation represents the line that passes through the points (-3, 7) and (9,-1)? oy--3x+5 o y=-x-7 o y=zx-7 oy - 12 2x+5

nekit [7.7K]

Answer:

use the slope equation (y2-y1) / (x2 -x1). plug in values and solve

(-1 -7) / (9 -(-3)) (subtracting a negative is the same as adding a positive)

-8 / 12 (simplify)

m = -2/3

then, plug in one of the points and the slope into the slope-point equation.

(y - y1) = m (x - x1) (the point-slope form)

y - 7 = -2/3( x - (-3)) (plug in the slope and point 1 values, then solve)

y - 7 = -2/3x - 2 (add 7 to both sides)

y = 2/3x +5 (the answer)

Step-by-step explanation:

I hope this helps :))

4 0

3 years ago

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f

Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.

The formula for the interval is:

p'₁± $Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }$

$Z_{1-\alpha /2}= Z_{0.95}= 1.648$

0.37±1.648* $\sqrt{\frac{0.37*0.63}{1000} }$

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂± $Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }$

0.48±1.648* $\sqrt{\frac{0.48*0.52}{1000} }$

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: $\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015$

Seniors: $\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016$

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0

3 years ago