Manny baked 48 cookies for a bake sale. 75% of the cookies were chocolate chip. How many cookies were NOT chocolate chip

Please help me with this one question. THANKS!

geniusboy [140]

The answer is A y-28=3(x-6)

4 0

3 years ago

Find the value of 10! (10-3) A. 70 B. 120 O C. 5040 O D. 720

GarryVolchara [31]

Answer:

The answer is option D.

Step-by-step explanation:

$\frac{10!}{(10 - 3) ! } = \frac{10!}{7!} = \frac{3628800}{5040} = 720$

Hope this helps you

8 0

3 years ago

What is the center of the circle that you can circumscribe about a triangle with vertices A(2, 6), B(2, 0), and C(10, 0)?

denis23 [38]

The three points A,B,C are all points on this circle.
Each point is then equal distance from the center, that distance being the radius of the circle.
Using the distance formula, we can find the center of the circle (x,y):
$d^2 = (x-x_0)^2 + (y-y_0)^2$
Plugging in points A and B into distance formula, then setting them equal to each other gives:
$(x-2)^2+(y-6)^2 = (x-2)^2 + y^2$
Right away we can cancel out the x terms leaving:
$(y-6)^2 = y^2$
Expand Left side and Solve for y:
$y^2 -12y +36 = y^2$
$y = 3$
Plug in points B and C as before:
$(x-2)^2 + y^2 = (x-10)^2 + y^2$
Here we can cancel the y-terms.
Expand and solve for x:
$x^2 -4x+4 = x^2 -20x+100$
$16x = 96$
$x = 6$
Therefore the center of the circle is the point (6,3)

6 0

3 years ago

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 434 gram setting. It is

yan [13]

Answer:

No, there is not sufficient evidence to support the claim that the bags are underfilled

Step-by-step explanation:

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 434 gram setting.

This means that the null hypothesis is:

$H_{0}: \mu = 434$

It is believed that the machine is underfilling the bags.

This means that the alternate hypothesis is:

$H_{a}: \mu < 434$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

434 is tested at the null hypothesis:

This means that $\mu = 434$

A 9 bag sample had a mean of 431 grams with a variance of 144.

This means that $X = 431, n = 9, \sigma = \sqrt{144} = 12$

Value of the test-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{431 - 434}{\frac{12}{\sqrt{9}}}$

$z = -0.75$

P-value of the test:

The pvalue of the test is the pvalue of z = -0.75, which is 0.2266

0.2266 > 0.01, which means that there is not sufficient evidence to support the claim that the bags are underfilled.

7 0

3 years ago

Could you help asap! Brainliest and 50 points!

olga2289 [7]

(5x^2-21x-20)/(5x^2-16x-16)

((x-5)(5x+4))/((x-4)(5x+4))

(x-5)/(x-4)

x² - 3 x - 54 = x² + 6 x - 9 x - 54 = x ( x + 6 ) - 9 ( x + 6 ) = ( x + 6 ) ( x - 9 )

x² - 18 x + 81 = ( x - 9 )²

x² + 12 x + 36 = ( x + 6 )²

... = ( x + 6 ) · ( x - 9 ) / ( x - 9 )² * ( x + 6 )² / ( x + 6 ) = ( after cancellation )

= ( x + 6 )² / ( x - 9 )

After solving the given problem, I’ve been able to get 3(x-2) / 2(x-3) as the simplified form of the given equation. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

If I understand clearly, the expression is

(18st^4/52s^3t)(16s^3/9s^2t)

To simply the expression, we simply have to multiply the constants and reduce the product to the lowest terms and apply the laws on exponents to simplify the variables. So, the answer would be:

8t^2/13s

(40v^2)/(35v^4) / (20v^3)/(5v)

(8)/(7v^2) / (4v^2/1)

(8)/(7v^2) * (1 / 4v^2)

2/(7v^4)

3 0

3 years ago

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