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1 year ago

This legend about Admiral Nelson, like other

Mathematics

1 answer:

il63 [147K]1 year ago

6 0

Correct option is E, like legends about other naval heroes, is

It does this by contrasting "this legend" with "legends about other naval heroes," so avoiding the mistake of the original. There is an unreasonable comparison in option (A). "This legend" is contrasted with "other naval heroes."

There is an agreement mistake in option (B). For the single subject "this legend," the plural verb "are" is used. There is an unreasonable comparison in option (C). "This legend" is contrasted with "other naval heroes." Choice (D) has an agreement mistake.

For the single subject "this legend," the plural verb "are" is used. Option E can be used to rewrite the sentence's highlighted portion. By contrasting "this legend" with "legends about other naval heroes," the inaccuracy is fixed. As a result, choice E is the right answer.

Here's a question with an answer similar to this Admiral Nelson:

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15 Points!

Xelga [282]

$f(x)=\dfrac{x-5}{x^2-1}\\\\\text{the horizontal asymptotes}\\\\x^2-1=0\to x^2=1\to x=\pm\sqrt1\to x=-1\ \wedge\ x=1\\\\\text{the vertical asymptotes}\\\\y=\lim\limits_{x\to\pm\infty}\dfrac{x-5}{x^2-1}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x}{x^2}+\frac{5}{x^2}}{\frac{x^2}{x^2}-\frac{1}{x^2}}=\dfrac{0+0}{1-0}=0$

Answer: <span>B) x = 1, x = -1, y = 0</span>

8 0

2 years ago

Suppose that Upper X has a discrete uniform distribution f left-parenthesis x right-parenthesis equals StartLayout left-brace1st

emmasim [6.3K]

Answer:

the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

Step-by-step explanation:

Given the data in the question;

x f(x) xp(x) x²p(x)

1 1/3 0.33333 0.33333

2 1/3 0.66667 1.33333

3 1/3 1.00000 3.0000

∑ 2.0000 4.6667

∑(xp(x)) = 2

∑(x²p(x)) = 4.6667

Variance σ² = ∑(x²) - ∑(x)² = 4.6667 - (2)² = 4.6667 - 4 = 0.6667

standard deviation σ = √variance = √0.6667 = 0.8165

Now since, n = 33 which is greater than 30, we can use normal approximation

for normal distribution z score ( x-μ)/σ

mean μ = 2

standard deviation = 0.817

sample size n = 33

standard of error σₓ = σ/√n = 0.817/√33 = 0.1422

so probability will be;

p( 2.1 < X < 2.4 ) = p(( 2.1-2)/0.1422) < x"-μ/σₓ < p(( 2.4-2)/0.1422)

= 0.70 < Z < 2.81

= 1 - ( 0.703 < Z < 2.812 )

FROM Z-SC0RE TABLE

= 1 - ( 0.25804 + 0.49752 )

= 1 - 0.75556

p( 2.1 < X < 2.4 ) = 0.2444

Therefore, the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

7 0

3 years ago

What is the pattern of odd and even products in each row

andrew11 [14]

There isn't a picture of a pattern to go by.

5 0

2 years ago

1. A-Tunes requires a membership fee of $12 a month and $1 for each download.

matrenka [14]

<h3>Answer:</h3>

After 16 downloads the charges will equal each other.

<h3>Step-by-step explanation:</h3>

First set up an equation

x = Downloads

12 + 1x = 5 + 1.5x

Solve

12 + 1x = 5 + 1.5x <- Rewrite your equation

-5 -5 <- Subtract 5 from both sides

8 + 1x = 1.5x <- Simplify

-1x -1x <- Subtract 1 from both sides

8 = 0.5x <- Simplify

8 = 1/2x <- Convert 0.5 to a fraction to make the next step easier

16 = x <- Multiply both sides by 2 to divide by

a fraction

4 0

3 years ago

8th GRADE MATH

s2008m [1.1K]

The answer for the question is a

6 0

2 years ago

Read 2 more answers