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3 years ago

11

Larry's evening is spent reading and watching TV. He can spend no more than 4 hours total on these activities, and he wants to s

pend more time reading than watching TV. Which graph represents this situation?

2 answers:

maks197457 [2]3 years ago

8 0

Please provided the graphs to choose from.

barxatty [35]3 years ago

3 0

Answer:

where is the graph

Step-by-step explanation:

You might be interested in

Check my answer?

Effectus [21]

D = 2r => 12.6 = 2*r => r=6.3
A = π*r^2 = (3.14) * (6.3)^2 = (3.14)*(39.69) = 124.6266 which is approximately 124.63

Hope this helps!

6 0

3 years ago

Read 2 more answers

Of the sundaes recently sold at Ice Cream Haven, 4 had nuts and 20 did not. Considering this

Svetlanka [38]

4/24= 1/6
(1/6)*84= 14 had nuts

20/24= 5/6
(5/6)* 84 = 70 no nuts

4 0

3 years ago

Simplify -3w-5(4x-4w)-4x<br>

dem82 [27]

-3w-5(4x-4w)-4x

multiply the bracket by -5

(-5)(4x)=-20x

(-5)(-4w)=20w

-3w-20x+20w-4x

-3w+20w-20x-4x ( combine like terms)

answer:

17w-24x or -24x+17w

5 0

2 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by:

$z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

4 0

3 years ago

Drag the phrases into the box to match each expression.

JulijaS [17]

Answer:

3 /4 x+ 1 /2 one-half more than three-fourths of a number

3/ 4 − 1 /2 x three-fourths minus one-half of a number

3/ 4 −(x+ 1/ 2 ) three-fourths minus the sum of a number and one-half

7 0

3 years ago