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3 years ago

Is this a quarter or not because im getting confused

Mathematics

2 answers:

kkurt [141]3 years ago

6 0

It’s a nickel a quarter would at least have a eagle or Washington on one side a nickel has Jefferson on it which shows on the picture. So it’s a nickel not a quarter

Komok [63]3 years ago

4 0

yes it is Ur welcome

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If g(x)=x²-2 and p(x)=2x+3, find (g·p)(x)

xxMikexx [17]

2X^3+3x^2-4x-6

You do x^2-2 times 2x+3 and use FOIL method to get answer

4 0

3 years ago

Enter an equation in standard form for the line.<br> Slope is -4, and (-9, -12) is on the line.

AfilCa [17]

Answer:

$4x + y +48=0$

Step-by-step explanation:

We are here given the slope of the line and one point through which it passes . The slope of the line is -4 and the point is (-9,-12) .

Here we can use the point slope form of the line as ,

$\sf\longrightarrow y-y_1 = m(x-x_1)$

Substitute the respective values ,

$\sf\longrightarrow y - (-12) = -4\{ x -(-9)\}$

Simplify LHS and RHS ,

$\sf\longrightarrow y + 12 = -4( x +9)$

Open the brackets in RHS ,

$\sf\longrightarrow y + 12 = -4x -36$

Add 36 and 4x to both sides ,

$\sf\longrightarrow \underline{\boxed{\orange{\sf 4x + y + 48 =0}}}$

8 0

3 years ago

Determine whether the following individual events are independent or dependent. Then find the probability of the combined event.

VashaNatasha [74]

Answer:

Here we have four individual events.

First selection, second selection, third selection and fourth selection.

Where initially, the options are 16 Americans and 13 Canadians.

Now, to see if the events are dependent or independent.

Suppose that in the first selection, an American is selected.

Then the second selection has 15 Americans and 13 Canadians as options.

Now, if in the first selection, a Canadian is selected.

Then in the second selection, the options will be 16 Americans and 12 Canadians.

Then the events are not independent.

Now we want to find the probability of randomly selecting a four-person committee consisting entirely of Canadians from a pool of 16 Americans and 13 Canadians

The probability of randomly selecting a Canadian is equal to the quotient between the number of Canadians, and the total number of postulants.

For the first selection, we have 13 Canadians, and 29 people in total, then the probability is:

P1 = 13/29.

Suppose that a Canadian is selected, now there are 12 Canadians and 28 people in total, then the probability for the second selection is:

P2 = 12/28

with the same reasoning as above, the probabilities for the third and fourth selections will be:

P3 = 11/27

P4 = 10/26.

The joint probability will be equal to the product of the individual probabilities, this is:

P = P1*P2*P3*P4 = (13/29)*(12/28)*(11/27)*(10/26) = 0.03

3 0

3 years ago

In the 2009 General Social Survey, respondents were asked if they favored or opposed death penalty for people convicted of murde

sleet_krkn [62]

Answer:

The calculated 99% confidence interval is wider than the 95% confidence interval.

Step-by-step explanation:

We are given the following in the question:

95% confidence interval for the population proportion

(0.65, 0.69)

Let $\hat{p}$ be the sample proportion

Confidence interval:

$p \pm z_{stat}(\text{Standard error})$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Let x be the standard error, then, we can write

$\hat{p} - 1.96x = 0.65\\\hat{p}+1.96x = 0.69$

Solving the two equations, we get,

$2\hat{p} = 0.65 + 0.69\\\\\hat{p} = \dfrac{1.34}{2} = 0.67\\\\x = \dfrac{0.69 - 0.67}{1.96} \approx 0.01$

99% Confidence interval:

$p \pm z_{stat}(\text{Standard error})$

$z_{critical}\text{ at}~\alpha_{0.01} = 2.58$

Putting values, we get,

$0.67 \pm 2.58(0.01)\\=0.67 \pm 0.0258\\=(0.6442,0.6958)$

Thus, the calculated 99% confidence interval is wider than the 95% confidence interval .

4 0

3 years ago

I need help on this one please and thank you

OleMash [197]

You buy ten packs of hot dogs. you buy six packs of buns. you buy sixty hot dogs. and now you've got diabetes

6 0

4 years ago