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3 years ago

Which of the following options is an odd function?

Mathematics

1 answer:

xxMikexx [17]3 years ago

3 0

Answer: The answer is C

Step-by-step explanation:

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Write an equation for the n th term of the arthemeric sequence 6,3,0,-3

s344n2d4d5 [400]

Answer:

x nth term =6–3(n-1)

Step-by-step explanation:

Since they start at 6 its X nth term n=6- but since the since one is 6 its 3(n-1) so x nth term=6-3(n-1)

3 0

3 years ago

Simplify completely. 10 points (:

zubka84 [21]

Answer:

2y^4√5

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

hope i helped :) !!!!!!!

brainliest !??!

8 0

3 years ago

Read 2 more answers

(1 pt) Which graphs show the solution to the given inequalities? Column A Column B 1. 2. 3. 4. A. Number line with positive and

ladessa [460]

I don't know but its at the tip of my tounge

4 0

2 years ago

The relative locations of Pedro's house, the grocery store, and the bank are shown in the diagram.

jolli1 [7]

Answer:

9.19 mi. is the distance between Bank and Grocery store.

Step-by-step explanation:

As we know in a given triangle with sides a, b, c which are the opposite sides of ∠A, ∠B, ∠C, follow the property

$\frac{a}{sinA} = \frac{b}{sinB}=\frac{c}{sinC}$

Now we apply this property in the given triangle.

Let the distance between Grocery store and Bank = x mi

$\frac{11}{sin75}=\frac{x}{sin54}$

$x=\frac{(sin54)(11)}{sin75}$

$\frac{(.81)(11)}{.97}=9.19$

So the answer is 9.19 mi.

4 0

3 years ago

Read 2 more answers

A market research company conducted a survey to find the level of affluence in a city. They defined the category "affluence" for

malfutka [58]

Answer:

A 95% confidence interval for this population proportion is [0.081, 0.159].

Step-by-step explanation:

We are given that a market research company conducted a survey to find the level of affluence in a city.

Out of 267 persons who replied to their survey, 32 are considered affluent.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who are considered affluent = $\frac{32}{267}$ = 0.12

n = sample of persons = 267

p = population proportion

Here for constructing a 95% confidence interval we have used One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.12-1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ , $0.12+1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ ]

= [0.081, 0.159]

Therefore, a 95% confidence interval for this population proportion is [0.081, 0.159].

4 0

3 years ago