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3 years ago

Is the number rational or irrational?

1 answer:

6 0

1. irrational: √48 = 6.92820323028

2. rational: √9 = 3

3. irrational: √39 = 6.2449979984

4. rational: √289 = 14

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In an August 2012 Gallup survey of 1,012 randomly selected U.S. adults (age 18 and over), 536 said that they were dissatisfied w

soldier1979 [14.2K]

Answer:

Yes, result is significant ; PVALUE < α

Step-by-step explanation:

Given :

x = 536

n = sample size = 1012

Phat = x / n = 536 / 1012 = 0.5296 = 0.53

H0 : P0 = 0.5

H1 : P0 > 0.5

Test statistic :

(Phat - P0) ÷ sqrt[(P0(1 - P0)) / n]

1-P0 = 1 - 0.5 = 0.5

(0.53 - 0.5) ÷ sqrt[(0.5*0.5)/1012]

0.03 ÷ 0.0157173

= 1.9087

Pvalue :

Using the Pvalue from test statistic :

Pvalue = 0.02815

To test if result is significant :

α = 0.05

0.02815 < 0.05

Pvalue < α ; Hence, result is significant at α=0.05; Hence, we reject H0.

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3 years ago

Please help asap! if you could also explain how you did it so I can answer the other questions by myself

Usimov [2.4K]

Answer:

Step-by-step explanation:

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3 years ago

Four gallons of gas cost $16.80. what is the unit rate?

SashulF [63]

1 gallon is $4.20
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2 years ago

What is the position of 7 in the number 876,543?

Butoxors [25]

A. The ten-thousands place.
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2 years ago

Read 2 more answers

Please help

Alekssandra [29.7K]

Answer:

25 in x 15 in

Step-by-step explanation:

Given:

Length = 3/5 the width
Area = 375 in²

Let width = $x$

Therefore, length = 3/5 $x$

First create an equation for the area of the picture based on the given information for its width and length:

$\begin{aligned} \implies \textsf{Area of original picture} & = \sf width \times length\\& = x\left(\dfrac{3}{5}x\right)\\& = \dfrac{3}{5}x^2\end{aligned}$

We are told the area of the enlarged picture is 375 in². Therefore, substitute this into the equation and solve for $x$ to find the width of the enlarged picture:

$\begin{aligned}\textsf{Area} & = 375\\ \implies \dfrac{3}{5}x^2 & = 375\\ x^2 & =375 \cdot \dfrac{5}{3}\\ x^2 & =625\\ x & = \sqrt{625}\\ x& = 25\end{aligned}$

Therefore, the width of the enlarged picture is 25 in.

Substitute the found value of $x$ into the expression for length to find the length of the enlarged picture:

$\begin{aligned}\sf Length & = \dfrac{3}{5}x\\\\\implies \sf Length & = \dfrac{3}{5}(25)\\\\& = 15\: \sf in\end{aligned}$

Therefore, the dimensions of the enlarged picture are <u>25 in x 15 in</u>. The width is 25 in and the length is 15 in, as the length is 3/5 of the width.

5 0

2 years ago