Kids go cart with a mass of 40 kg has a momentum of 800 kg x m/s. What speed is the go cart traveling?

Anton [14]

Answer:

I'm sure it is 180 degrees

8 0

2 years ago

A trapezoid has an area of 780 square meters. The height of the trapezoid is 40 meters, and the length of the longer base is tw

o-na [289]

Answer:

idk I'm just here for points sorry bro

6 0

2 years ago

I need help I don’t know this question

murzikaleks [220]

So singlehandedly, November has 25% of the total rain in one year of New York's, which is 62.51 inches.

So, November receives: *percentage* x *total*
= 25% x 62.51 = 15.6275 in

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So, in November, New Yorks receives 25% of total rain in one year, which is 15.6275 inches.
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I'd like my Brainliest pls ;W;

4 0

2 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

1 year ago

PLEASE HELP

zhuklara [117]

Beck add like terms so you get
4x^2 -4x-15
lucy factor our negative so you left with -4x^2 +5x -10 +8x^2 -9x -5
simplify 4x^2 -4x -15
therefore both students answers simplify to 4x^2 -4x -15

7 0

2 years ago