-12 = 6+ 2/3 y Pleaseeeee help me ❤️

28 cookies out of 40 cookies are chocolate chip. What percent of the cookies are chocolate chip?

omeli [17]

Answer:

I’m pretty sure the answer would be 64%

7 0

2 years ago

Read 2 more answers

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

3 years ago

Two is subtracted from a number, and then the difference is multiplied by 5. The result is 30

Vanyuwa [196]

Answer:

8

Step-by-step explanation:

8-2=6

6x5=30

6 0

3 years ago

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Can't figure it out!!

Zanzabum

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)= \cfrac{1}{x} \qquad 
\begin{cases}
x_1=2\\
x_2=b
\end{cases}\implies \cfrac{f(b)-f(2)}{b-2}=-\cfrac{1}{10}
\\\\\\
\cfrac{\frac{1}{b}-\frac{1}{2}}{b-2}=-\cfrac{1}{10}\implies 
\cfrac{\frac{2-b}{2b}}{b-2}=-\cfrac{1}{10}\implies \cfrac{2-b}{2b}=-\cfrac{b-2}{10}
\\\\\\
$

$\bf \cfrac{2-b}{2b}=\cfrac{2-b}{10}\implies 20-10b=4b-2b^2\impliedby cross-multiplying
\\\\\\
2b^2-4b-10b+20=0\implies 2b^2-14b+20=0
\\\\\\
b^2-7b+10=0\implies 
(b-2)(b-5)=0\implies b=
\begin{cases}
\boxed{5}\\
2
\end{cases}$

8 0

3 years ago

Shira’s Shoes sold 875,000 pairs of sandals in June, which was 70% of the total number of shoes sold. How many shoes did the com

azamat

Answer:

1250000

Step-by-step explanation:

Let number of the total number of shoes sold be 'y'

y*(70/100)= 875000

70y = 875000*100

y = (875000*100)/70

y = 1250000

3 0

3 years ago

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