Someone plz help me out i need to pass

Brainliest!!! just help me with one question!! :)

geniusboy [140]

C, 5 slices I believe!

5 0

3 years ago

Read 2 more answers

X^2 + Y^2 +ax + 2y + 3 = 0<br><br> What number is this equation the equation of a circle?

Sav [38]

Answer:

Eq: (x+a/2)²+(y+1)²=(a²-8)/4

Center: O(-a/2, -1)

Radius: r=0.5×sqrt(a²-8)

Mandatory: a>2×sqrt(2)

Step-by-step explanation:

The circle with center in O(xo,yo) and radius r has the equation:

(x-xo)²+(y-yo)²=r²

We have:

x²+y²+ax+2y+3=0

But: x²+ax=x²+2(a/2)x+a²/4-a²/4= (x+a/2)²-a²/4

And

y²+2y+3=y²+2y+1+2=(y+1)²+2

Replacing, we get:

(x+a/2)²-a²/4+(y+1)²+2=0

(x+a/2)²+(y+1)²=a²/4-2=(a²-8)/4

By visual inspection we note that:

- center of circle: O(-a/2, -1)

- radius: r=sqrt((a²-8)/4)=0.5×sqrt(a²-8). This means a²>8 or a>2×sqrt(2)

6 0

3 years ago

Pam needs to spend less than $40 on her school supplies. She spends $12 on a

Cloud [144]

Answer:

28

Step-by-step explanation:

Total amount of money she has=$40

Spends=12

40-12=28

She has $28 dollars left to spend on her binders.

3 0

3 years ago

Read 2 more answers

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

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

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

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.