Distance from the x-axis is 2

PLSS HELP ME WITH THIS! 15 POINTS!

irina [24]

Answer:

2w - 13

Step-by-step explanation:

here we just simplify the expression,

3w - 4 - w - 9

becomes:

2w - 13

5 0

1 year ago

Read 2 more answers

You want to estimate the mean amount of time college students spend on the Internet each month. How many college students must y

KonstantinChe [14]

Answer:

752.95

Step-by-step explanation:

Data provided in the question

The standard deviation of population = 210

The Margin of error = 15

The confidence level is 75%, so the z value is 1.96

Now the required sample size is

$= 1.96^2\times \frac{210^2}{15^2}$

= 752.95

Hence, the number of college students spends on the internet each month is 752.95

Simply we considered the above values so that the n could come

8 0

2 years ago

Divide 794.1 by 7.61 Point round off your answer to two decimal places

dolphi86 [110]

<span>By dividing the two numbers we get 104.3495... Rounding this number to two decimal places means that we must keep just 2 digits after the "." symbol, and round the last one of them. In our case, 104.3495... becomes 104.35.</span>

5 0

3 years ago

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the

Sergio039 [100]

Answer:

The estimate is $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is $\r p__{hat}} = 0.37$

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

$\alpha = 100 - 90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=> $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=> $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

$P__{hat}} \pm E$

substituting values

$0.37 \pm 0.0348$

5 0

2 years ago

A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft

lakkis [162]

$\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}$

$\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}$

8 0

2 years ago