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klasskru [66]
3 years ago
13

What is the slope of a line that is parallel to y equals 3x + 5

Mathematics
2 answers:
Sergio039 [100]3 years ago
6 0

A line parallel to this one would have a slope of 3. where m is the slope and b is the y intercept. In this case, the equation y=3x+5 is already in slope intercept form, which means the slope is 3.

sweet-ann [11.9K]3 years ago
5 0
A line parallel to this one would have a slope of 3. where m is the slope and b is the y intercept. In this case, the equation y=3x+5 is already in slope intercept form, which means the slope is 3.
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What is the probability of drawing a blue marble?<br><br> 3/7<br> 70%<br> 1/3<br> 3/10
vazorg [7]

Answer:

3/10

Step-by-step explanation:

There are 20 marbles.  6 of them are blue.

P = 6/20 = 3/10

4 0
3 years ago
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2 y - 8 / 3 equals a + 3 B / 4 solving for y
zhuklara [117]
-2x+5
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3 years ago
A small theater had 6 rows of 26 chairs each. An extra 9 chairs have just been brought in. How many chairs are in the theater no
Andrej [43]

Answer:165

Step-by-step explanation:

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3 years ago
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A restaurant will select 1 card from a bowl to win a free lunch, jimmy puts 10 cards in the bowl. The bowl has a total of 150 ca
Phantasy [73]

The odds of jimmy winning a free lunch is 0.066.

<h3>What is Probability ?</h3>

Probability is the likeliness of an event to happen.

It is given that

A restaurant will select 1 card from a bowl to win a free lunch,

jimmy puts 10 cards in the bowl.

The bowl has a total of 150 cards in it

The odds of jimmy winning a free lunch is given by

= No. of chances / Total Outcomes

= 10/150

= 1/15

= 0.066

Therefore the odds of jimmy winning a free lunch is 0.066.

To know more about Probability

brainly.com/question/11234923

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3 0
2 years ago
Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

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

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:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
4 years ago
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