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elena-14-01-66 [18.8K]

3 years ago

7

John was given a math equation. He completed it in 5 seconds. Is his answer correct? If it is, show how he got the answer. Show

your work. See picture down below.

2 answers:

steposvetlana [31]3 years ago

6 0

His answer is correct because;
1 bag in addition to 1 bag is 2 bags

lutik1710 [3]3 years ago

5 0

Answer:

His work is correct, easiest question in the world!LOL

Step-by-step explanation:

1+1= 2

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Give a big-o estimate for the number of operations (where an operation is an addition or a multiplication) used in this segment

Marianna [84]

We have a "rectangular" double loop, meaning that both loops go to completion.
So there are 3*4=12 executions of t:=t+ij.

Assuming two operatiions per execution of the innermost loop, (i.e. ignoring the implied additions in increment of subscripts), we have 12*2=24 operations in all.

Here the number of operations (+ or *) is exactly known (=24).

Big-O estimates are used for cases with a varying scale of operations, governed by a variable (usually n) to indicate the sensitivity of the number of operations relative to a change in the size of n.

Here we do not have a scale, nor n is defined. The number of operations is constant and known at 24. So a variable is required to find the big-O estimate.

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3 years ago

Please answer for brainliest<br> what is a word problem for<br><br><br> 5x + 10 ≥ 50

scoundrel [369]

Use this word problem just change the numbers

7 0

3 years ago

In a state's Pick 3 lottery game, you pay $1.15 to select a sequence of three digits (from 0 to 9), such as 966. If you sel

m_a_m_a [10]

Answer:

A) 1000 different selections

B) probability of winning = 0.001

C) Net profit on winning = $257.20

D) E(y) = -0. 89165

E) Option B - The Pick 4 game is a better bet because it has a larger expected value.

Step-by-step explanation:

A) For the different selections possible, since each of the digits from 0 - 9 can take a total of 10 digits, then the different selections possible will be = 10³ = 1000 different selections

B) To find the probability of winning, since only one pick can be a winning one. Thus probability of winning = 1/1000 = 0.001

C) Net profit on winning = Amount won - Amount staked

Amount won = $258.35

Amount staked = $1.15

Thus;

Net profit on winning = $258.35 - $1.15

Net profit on winning = $257.20

D) If we call the net profit "y", then Expected value is:

E(y) = 257.2(0.001) - 1.15(1 - 0.001)

E(y) = 0.2572 - 1.14885

E(y) = -0. 89165

E) We are told that:If you bet $ 1.15 in a certain state's Pick 4 game, the expected value is negative −$0.89

The expected value of -0.89 in pick 4 is better than that in pick 3 because it's larger.

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3 years ago

Substitute this equation y=-2x-12 if y=-8

lozanna [386]

Answer:

x=-2

Step-by-step explanation:

6 0

3 years ago

Suppose that a researcher is designing a survey to estimate the proportion of adults in your state who oppose a proposed law tha

irinina [24]

Answer:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

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

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Solution to the problem

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

The margin of error desired for this case is $ME= \pm 0.02$ equivalent to 2% points

For this case we need to assume a confidence level, let's assume 95%. And since we don't have prior estimation for the population proportion of interest the best value to do an approximation is $\hat p =0.5$

In order to find the critical value we need to take in count that we are finding the margin of error for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

Now we have all the values needed and if we replace into equation (b) we got:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

5 0

3 years ago