Describe the locations of 3 and -3 with respect to 0 on a number line.

Madison works as a salesperson at an electronics store and sells phones and phone accessories, and makes a fixed commission for

Jet001 [13]

Answer:

Price of each phone: $\$9$

Price of each accessory: $\$3.5$

Step-by-step explanation:

Let be "p" the price in dollars of each phone and "a" the price in dollars of each accessory.

Set up a system of equations:

$\left \{ {{8p+10a=107} \atop {4p+23a=116.50}} \right.$

You can use the Elimination Method to solve the system.

Multiply the second equation by -2 and add both equations:

$\left \{ {{8p+10a=107} \atop {-8p-46a=-233}} \right.\\............................\\-36a=-126\\\\a=\frac{-126}{-36}\\\\a=3.5$

Substitute the value of "a" into the first equation and solve for "p":

$8p+10(3.5)=107\\\\8p=107-35\\\\p=\frac{72}{8}\\\\p=9$

4 0

4 years ago

Jack spends 1 2/5 as long on his homework as Jill. Last week, Jill spent 6 3/4 hours doing homework. How long did Jack spend doi

-BARSIC- [3]

6 3/4 hours = (6*60)+45 = 405 minutes
7/5 = 1,4
405 minutes * 1,4 = 567 minutes
= Jack spent 9 3/4 hours

8 0

3 years ago

Which number line represents the solutions to [x - 5] = 1?

Strike441 [17]

Answer:

there should only be two points at 6 and 4

Step-by-step explanation:

[x-5]=1

x-5=+ 1 or -1

x-5= 1

x-5= -1

x=6 or 4

7 0

4 years ago

Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later (in 2008) that perc

Dima020 [189]

Answer:

a) $ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

b) $ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

Step-by-step explanation:

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

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 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

If we replace the values into equation (a) for 1983 we got:

$ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

Part b

Since is the same confidence level the z value it's the same.

If we replace the values into equation (a) for 2008 we got:

$ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

3 0

4 years ago

A factory sells backpacks for $40.00 each. The cost to make 1 backpack is $10.00. In addition to the cost of making backpacks, t

vlabodo [156]

Answer:

300 Backpacks each week]

Step-by-step explanation:

dont have an explanation srry but the answer is totally 300 backpacks a week

4 0

3 years ago

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