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2 years ago

im giving my friends 9 cupcakes thats 60 percent of my order how many more cupcakes do i need to get my order to 100 percent

Mathematics

1 answer:

Gwar [14]2 years ago

4 0

Answer:

6 more cupcakes.

Step-by-step explanation:

60% of 15 is 9, so 15-9 is 6.

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Divide the following 81x⁷yz³ by -3x²y²z²

Readme [11.4K]

Answer:

-27x^5y^-1 z

Step-by-step explanation:

-3x²y²z²

-27x^5y^-1 z

4 0

3 years ago

HELP ASAP!!!!!!!!!!!!

torisob [31]

Answer:

163 in²

Step-by-step explanation:

To find the area, I divided the shape into squares and rectangles. Then, I found the area of the squares and rectangles and added them,

Area = (7 x 7) + (18 x 5) + (6 x 4)
=> (49) + (90) + (24)
=> 163 in²

Hence, the area is 163 in.²

7 0

2 years ago

Eight parentheses x plus two parentheses

jarptica [38.1K]

Answer:

Step-by-step explanation:

8+2=10

3 0

3 years ago

The ages of 11 students enrolled in an on-line macroeconomics course are given in the following steam-and-leaf display:

blsea [12.9K]

Answer:

The standard deviation of the age distribution is 6.2899 years.

Step-by-step explanation:

The formula to compute the standard deviation is:

$SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}$

The data provided is:

X = {19, 19, 21, 25, 25, 28, 29, 30, 31, 32, 40}

Compute the mean of the data as follows:

$\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{x_{i}}$

$=\frac{1}{11}\times [19+19+21+...+40]\\\\=\frac{299}{11}\\\\=27.182$

Compute the standard deviation as follows:

$SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}$

$=\sqrt{\frac{1}{11-1}\times [(19-27.182)^{2}+(19-27.182)^{2}+...+(40-27.182)^{2}]}}\\\\=\sqrt{\frac{395.6364}{10}}\\\\=6.28996\\\\\approx 6.2899$

Thus, the standard deviation of the age distribution is 6.2899 years.

7 0

3 years ago

What is -3x^2-4x-4=0

ss7ja [257]

Answer:

<h2>no real solution</h2>

Step-by-step explanation:

$-3x^2-4x-4=0\qquad\text{change the signs}\\\\3x^2+4x+4=0\\\\\text{use the quadratic formula:}\\\\\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then an equation has two solutions:}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\text{We have}\ a=3,\ b=4,\ c=4.\\\\\text{Substitute:}\\\\b^2-4ac=4^2-4(3)(4)=16-48=-32$

8 0

3 years ago