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2 years ago

Regardless of the algorithm being used, a search through an array is always performed:

1 answer:

7 0

The answer to these is none, not when it is from the highest and lowest element, not even from lowest to highest element. And no, you actually can not use a binary element to find the answer. So the ptr is a pointer that points the variables that can address of the integer.

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I need help please don't know this

SashulF [63]

Answer:

9 is the right answer

Step-by-step explanation:

I. If c = - 4 and d = 10

then $\frac{1}{4}(-4^{3} + 10^{2} )$

= $\frac{1}{4}(-64 + 100)$

= $\frac{1}{4}(36)$

= 9

Hope that help :D

6 0

3 years ago

-5x-y= 8<br> -4x+y=2 solve the system

yan [13]

$\huge\fcolorbox{blue}{aqua}{Answer:}$

═════════════════════

1. $- 5x - y = 8$

<h3>Solve for x</h3>

$-5x-y=8$

$-5x=8+y$

$-5x=y+8$

$\frac{-5x}{-5}=\frac{y+8}{-5}$

$x=\frac{y+8}{-5}$

$x=\frac{-y-8}{5}$

<h3>Solve for y</h3>

$-5x-y=8$

$-y=8+5x$

$-y=5x+8$

$\frac{-y}{-1}=\frac{5x+8}{-1}$

$y=\frac{5x+8}{-1}$

$y=-5x-8$

────────────────────

2. $-4x-y=2$

<h3>Solve for x</h3>

$-4x-y=2$

$-4x=2+y$

$-4x=y+2$

$\frac{-4x}{-4}=\frac{y+2}{-4}$

$x=\frac{y+2}{-4}$

$x=-\frac{y}{4}-\frac{1}{2}$

<h3>Solve for y</h3>

$-4x-y=2$

$-y=2+4x$

$-y=4x+2$

$\frac{-y}{-1}=\frac{4x+2}{-1}$

$y=\frac{4x+2}{-1}$

$y=-4x-2$

════════════════════

Explanation:

#Carry on learning

7 0

2 years ago

Solve for slope. (-2,7) (-1,6)

Lostsunrise [7]

Answer:

m = - 1/3

Thats the answer.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma

nata0808 [166]

Answer:

a) $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c) $P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d) $P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e) $P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

$X \sim N(110000,40000)$

Where $\mu=110000$ and $\sigma=40000$

If we select a sample size of n =35 the standard error is given by:

$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

$P(X > 112000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

$P(X > 100000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

$P(100000$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

8 0

3 years ago

The answer to this question

WINSTONCH [101]

Answer:

Step-by-step explanation:

7 0

3 years ago