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7nadin3 [17]
3 years ago
9

Graph Equation 1: 6x - 3y = 3 Equation 2: 4x - 2y = 8 *

Mathematics
1 answer:
tatiyna3 years ago
7 0

6x - 3y = 3 will be graph 1

4x - 2y = 8 will be graph 2

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Wha it is 2,200 divided by 30 and 2,300 divided by 30
Paladinen [302]

Answer:

2,200/30=73.3333333

2,300/30=76.6666667

Step-by-step explanation:

6 0
3 years ago
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Algebra 2<br><br> Solve each equation to the nearest thousand <br><br> Help pls
borishaifa [10]
I don’t understand your question, can you give more details please
5 0
2 years ago
William has 8 1/4 cups of fruit juice.  If he divides the juice into 3/4 cup servings, how many servings will he have?
nikdorinn [45]

Answer: 11 servings

Step-by-step explanation:

To get the number of servings, all we need to do is divide 8 1/4 by 3/4

converting 8 1/4 to improper fraction will yield 33/4

8 1/4 ÷ 3/4

=33/4 ÷ 3/4

= 33/4 ÷ 4/3

The 4 at the numerator will cancel the 4 at the denominator. we will be left with;

=33/3 =11

Therefore, there will be 11 servings.

7 0
3 years ago
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How many numbers have absolute value 3?
kirill115 [55]

Answer:

how many numbers have absolute value 3? zero is your answer

how many numbers have absolute value a, if a>0? your answer is also zero

Step-by-step explanation:

4 0
3 years ago
There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f
Ann [662]

Answer:

0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647

0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

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{p(1-p)}{n}})

Solution to the problem

The estimated population proportion for this case is:

\hat p = \frac{350}{500}=0.7

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647

0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

7 0
2 years ago
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