Ann has seven balloons. Molly has a set less of balloons. How many balloons does molly have?

Mathematics

1 answer:

harina [27]2 years ago

4 0

0 balloons because if a set of balloons is 7 and Molly has a set less then it is 7-7=0

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(x-2)(x+2)(x²+4)-(x²+3)(x²-3)+2022

torisob [31]

The simplified form of the given expression is 2015

<h3>Simplifying expressions </h3>

The given expression is (x-2)(x+2)(x²+4)-(x²+3)(x²-3)+2022

The expression can be simplified as shown below

(x-2)(x+2)(x²+4) - (x²+3)(x²-3) + 2022

x(x+2)-2(x+2)(x²+4) - x²(x²-3) +3(x²-3) + 2022

(x²+2x-2x-4)(x²+4) - (x⁴-3x²+3x²-9) + 2022

(x²-4)(x²+4) - (x⁴-9) + 2022

x²(x²+4) -4(x²+4) - (x⁴-9) + 2022

(x⁴+4x²-4x²-16) - (x⁴-9) + 2022

(x⁴-16) - (x⁴-9) + 2022

x⁴ -16 -x⁴ +9 + 2022

x⁴ -x⁴ -16 +9 +2022

= 2015

Hence, the simplified expression is 2015

Learn more on Simplifying an expression here: brainly.com/question/723406

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3 0

1 year ago

Which is the ratio of the number of months that begin with the letter M to the total number of months in a year? 2 to 12 2 to 10

Damm [24]

March, may - 2 months
12 months in total

so it's 2 to 12

6 0

2 years ago

Read 2 more answers

The average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 oun

Artist 52 [7]

Answer:

The probability that the mean of this sample is less than 16.1 ounces of beverage is 0.0537.

Step-by-step explanation:

We are given that the average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 ounces.

A random sample of sixty-five 16-ounce beverage cans are selected

Let $\bar X$ = <u><em>sample mean amount of a beverage</em></u>

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean amount of a beverage = 16.18 ounces

$\sigma$ = standard deviation = 0.4 ounces

n = sample of 16-ounce beverage cans = 65

Now, the probability that the mean of this sample is less than 16.1 ounces of beverage is given by = P( $\bar X$ < 16.1 ounces)

P( $\bar X$ < 16.1 ounces) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{16.1-16.18}{\frac{0.4}{\sqrt{65} } }$ ) = P(Z < -1.61) = 1 - P(Z $\leq$ 1.61)