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3 years ago

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Two classes have a total of 50 students. One of the classes has 6 more students than the other. How many students are in the lar

ger class?

1 answer:

Naddik [55]3 years ago

8 0

Class 1= 6+x
class 2 = x
total= 50
so you need to find the x
50=6+x+x
50=2x+6
50-6=2x
44=2x
22=x
22+6
28 is the larger class

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Podría alguien ayudarme, es para mañana.

Ludmilka [50]

Answer:..$.$!:

Step-by-step explanation:

5 0

3 years ago

7x^2=9+x what are the values of x<br><br>Will give medal and points

DENIUS [597]

7x² = 9 + x Subtract x from both sides
7x² - x = 9 Subtract 9 from both sides
7x² - x - 9 = 0 Use the Quadratic Formula

a = 7 , b = -1 , c = -9

x = $\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ Plug in the a, b, and c values
x = $\frac{- (-1) \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Cancel out the double negative
x = $\frac{1 \pm \sqrt{(-1)^2 - 4(7)(-9)} }{2(7)}$ Square -1
x = $\frac{1 \pm \sqrt{1 - 4(7)(-9)} }{2(7)}$ Multiply 7 and -9
x = $\frac{1 \pm \sqrt{1 - 4(-63} }{2(7)}$ Multiply -4 and -63
x = $\frac{1 \pm \sqrt{1 + 252} }{2(7)}$ Multiply 2 and 7
x = $\frac{1 \pm \sqrt{1 + 252} }{14}$ Add 1 and 252
x = $\frac{1 \pm \sqrt{253} }{14}$ Split up the $\pm$
x = $\left \{ {{ \frac{1 + \sqrt{253} }{14} } \atop { \frac{1 - \sqrt{253} }{14} }} \right.$
The approximate square root of 253 is <span>15.905973.
</span>x ≈ $\left \{ { \frac{1 + 15.905973}{14} } \atop { \frac{1 - 15.905973}{14} }} \right$ Add and subtract
x ≈ $\left \{ {{ \frac{16.905973}{14} } \atop { \frac{14.905973}{14} }} \right.$ Divide
x ≈ $\left \{ {{1.2075} \atop {1.0647}} \right.$ Round to the nearest hundredth
x ≈ $\left \{ {{1.21} \atop {1.06}} \right.$

<span>
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7 0

3 years ago

Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was

andrezito [222]

Answer:

The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 100, p = 0.42$

92% confidence level

So $\alpha = 0.08$ , z is the value of Z that has a pvalue of $1 - \frac{0.08}{2} = 0.96$ , so $Z = 1.75$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.3336$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.5064$

The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

4 0

3 years ago

What is the answer to the equation -2x-8=-10

Oxana [17]

You need to get "x" by itself on one side

$-2x-8=-10$

Add positive eight on both sides---because its the opposite of negaitve

$-2x-8+8=-10+8$

$-2x=-2$

Divide -2 by -2

-2÷-2=1

<u> $x=1$ </u>

6 0

3 years ago

Help help help help help

Arada [10]

Answer:

12f-5f²+6f+10f²= 10f²-5f²+12f+6f=5f²+18f

5 0

3 years ago