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4 years ago

Rewrite the function W= 50+2.3(h-60) in order to express height as a function of weight

1 answer:

5 0

You need to solve the equation for h.

W= 50+2.3(h-60)

Distribute the 2.3.

W = 50 + 2.3h - 138

W = 2.3h - 88

Add 88 to both sides.

W + 88 = 2.3h

Switch sides.

2.3h = W + 88

Divide both sides by 2.3.

h = (W + 88)/2.3

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Solve this |2x-2|+6b

Anna71 [15]

Answer:

Step-by-step explanation:

its not letting me put the resolts but you can use this link and i hope it helps you :)

https://mathsolver.microsoft.com/en/solve-problem/%7C%202x-2%20%20%7C%20%20%2B6

6 0

3 years ago

60 points

Agata [3.3K]

Answer:

blank 1: 2xy(4x-5y)+3(4x-5y)

blank 2: (2xy+3)(4x-5y)

Step-by-step explanation:

Using factoring group to factor the polynomial 8x²y−10xy²+12x−15y

Step 1: Group the polynomial function

(8x²y−10xy²)+(12x−15y)

Step 2: Factor the common value and variables from the functions in parentheses.

= (2xy*4x - 2xy(5y)) + (3(4x) - 3(5y))

= 2xy(4x-5y)+3(4x-5y)

Step 3: Rearrange the expression by selecting one of the function in parentheses and combine those not in parenthesis.

(2xy+3)(4x-5y)

The final expression gives the factor form of the polynomial.

Hence the statement for each blank that correctly completes Lana's work are;

blank 1: 2xy(4x-5y)+3(4x-5y)

blank 2: (2xy+3)(4x-5y)

6 0

3 years ago

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

IRINA_888 [86]

Answer:

[0.5235, 0.5765]

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

b) 1,843 American adults

Step-by-step explanation:

The 99% confidence interval is given by

$\bf p\pm z^*\sqrt{p(1-p)/n}$

where

p = the proportion of American adults surveyed who said they have watched digitally streamed TV programming on some type of device = 55% = 0.55

$\bf z^*$ the z-score for a 99% confidence level associated with the Normal distribution N(0,1). We can do this given that the sample size (2,341) is big enough

n = sample size = 2,341

We can find the $\bf z^*$ value either with a table or with a spreadsheet.

In Excel use NORM.INV(0.995,0,1)

In OpenOffice Calc use NORMINV(0.995;0;1)

We get a value of $\bf z*$ = 2.576

and our 99% confidence interval is

$\bf 0.55\pm 2.576\sqrt{0.55*0.45/2341}=0.55\pm 2.576*0.0103=0.55\pm 0.265 = [0.5235, 0.5765]$

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

We are 99% confident that this interval contains the true population proportion.

(b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p?? (Round your answer up to the nearest integer.)

The sample size n in a simple random sampling is given by

$\bf n=\frac{(z^*)^2p(1-p)}{e^2}$