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Naddik [55]
3 years ago
12

A bag contains balls in a radio of 3 5 4 if there are twelve balls how many balls are in the bag!!

Mathematics
2 answers:
Sonja [21]3 years ago
6 0
Twelve balls are in the bag
Elena-2011 [213]3 years ago
4 0
Twelve balls are in the bag
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Which one would be the number line ???????
8090 [49]

Answer:

B

Step-by-step explanation:

1/2x - 2 > 0

collect like terms

1/2x > 0 + 2

1/2x > 2

Divide both sides by 1/2

x > 2 divided by 1/2

inverse of 1/2 is 2/1

x > 2 times 2/1

x > 4

8 0
2 years ago
Read 2 more answers
What is the equation of (4,10),(6,11) in standard form?​
SIZIF [17.4K]

ANSWER

The standard form is:

x - 2y =  - 16

EXPLANATION

The given line passes through (4,10)and (6,11).

The slope of this line is

m =  \frac{11 - 10}{6 - 4}

m =  \frac{1}{2}

The equation is given by the formula,

y-y_1=m(x-x_1)

We plug in the first point and the slope to get.

y-10= \frac{1}{2} (x-4)

2y - 20 = x - 4

x - 2y  =  - 20 + 4

x - 2y =  - 16

The standard form is

x - 2y =  - 16

7 0
3 years ago
Bernice estimated that 69% of 242 is 168. Is her estimate reasonable?
ycow [4]
Her estimate is not reasonable.

Since she estimated 69%, you could do either of these options. You could...
A. Divide 242 (or .5) to show how much half off would be, then compare her estimate with half of 242 (121). OR...
B. Simply divide 242 by 69% (.69) and compaor her estimate to the exact amount (166.98).

I hope this helps!
6 0
3 years ago
Which of the following is not a possible r-value -0.3 -1.6 0.8 0.3
Sliva [168]

Answer: -1.6 is not a possible r-value.

Step-by-step explanation:

R-values range from -1 to +1. Out of all these numbers, -1.6 does not belong. Hence, -1.6 is not a r-value.

3 0
3 years ago
Please help! I don't understand how to solve this problem
Ilia_Sergeevich [38]

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

The margin of error is:

M = z\frac{\sigma}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • \sigma is the standard deviation for the population.

Assuming an uniform distribution, the standard deviation is given by:

S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

M = z\frac{\sigma}{\sqrt{n}}

0.006 = 1.96\frac{1.1547}{\sqrt{n}}

0.006\sqrt{n} = 1.96 \times 1.1547

\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}

(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2

n =  142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

8 0
1 year ago
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