Rewrite the sum of 96 + 80 using the distrubutive property

2 answers:

6 0

96+80 is just 176 but we need to do the distributive properties. So distributive properties for example at a(b+c) but we need ours to equal 176. Let’s pick an a value, let’s say 4 so 4(b+c)=176. A quarter of 176 is 44, so b+c =44. B has to stay as a variable. Let’s say b= 30. C would be 14. So 4(b+14) would be a distributive property for 96+80

MissTica2 years ago

3 0

Yes I have the same problem with my phone I think it’s 190

You might be interested in

!!!!!!!!!!!!~~math is hard i could use some help i also raised the points and will mark brainliest uwu ~~!!!!!!!!!!!!!!

elena-s [515]

I think (jk) (x) means j times k

(jk) (x) = -7x(9 - 2x) //Using Distributive Property.

(jk) (x) = -63x + 14x^2 = 14x^2 - 63

//Hope it helps.

7 0

3 years ago

The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction

Len [333]

Answer:

(a) n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

For a sample size of n = 20

$P(\mu-200$

$=P(-0.37$

For a sample size of n = 50

$P(\mu-200$

$=P(-0.59$

For a sample size of n = 100

$P(\mu-200$

$=P(-0.83$

For a sample size of n = 500

$P(\mu-200$

$=P(-1.86$

n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ( $\bar x$ ) approaches the whole population mean ( $\mu_{x}$ ).