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Alex17521 [72]
3 years ago
6

Number 2 please help

Mathematics
1 answer:
Sveta_85 [38]3 years ago
5 0

Answer:

you just have to check the first and the third box

Step-by-step explanation:

put each into slop intercept form

y+6=-5(x-2) is the same as y=-5x+4

y-2=-5(x+6) is the same as y=-5x-28

y=-5x+4

y=-5x-28

all of them have a slope of -5, but you need to plug in (2,-6) to see which on is true

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Becky bought a magazine for $3.67, and the sales tax was 8.25%. How much sales tax did Becky pay on the magazine?
shutvik [7]
It would be A. or B

Hope This Helps!
4 0
3 years ago
Read 2 more answers
#1 Find the slope given the points (3,-20) and (5,8) *
Artyom0805 [142]

Hey there :)

To find the slope with <em>two points</em> we will need to use the equation  \frac{y^2-y^1}{x^2-x^1}

x^1    x^2       y^1  y^2

( 3 , -20 )  ( 5 , 8 )

Here is what the equation should look like when replaced with the numbers (<em>it will be a fraction</em>) :

                                     \frac{8 - (-20)}{5 - 3}

We know that <em>two negatives</em> cancel out and make a positive, so the numerator would actually be 8 + 20.

Solve the fraction and we get   \frac{28}{2}

28 <em>divided</em> by 2 is 14. The slope is 14.

6 0
3 years ago
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 335335 babies were​ born
strojnjashka [21]

Answer:

The 99​% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

Step-by-step explanation:

Confidence Interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 335, \pi = \frac{268}{335} = 0.8

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 - 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.7437

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 + 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.8563

For the percentage:

Multiplying the proportions by 100.

The 99​% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

7 0
3 years ago
Triangle ABC is a right triangle with mC = 90°, m B = 54°, and
rosijanka [135]

Answer:

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Step-by-step explanation:

YPU

5 0
3 years ago
Let Q(x, y) be the predicate "If x &lt; y then x2 &lt; y2," with domain for both x and y being R, the set of all real numbers.
Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is Q(x,y) is predicate that x then x^{2}. where x,y are rational numbers.

a)

when x=-2, y=1

Here -2 that is x  satisfied. Then

(-2)^{2}

4 this is wrong. since 4>1

That is x^{2}>y^{2} Thus Q(x,y) =Q(-2,1)is false.

b)

Assume Q(x,y)=Q(-5,2).

That is x=-5, y=2

Here -5 that is x this condition is satisfied.

Then

(-5)^{2}

25 this is not true. since 25>4.

This is similar to the truth value of part (a).

Since in both x satisfied and x^{2} >y^{2} for both the points.

c)

if Q(x,y)=Q(3,8) that is x=3 and y=8

Here 3 this satisfies the condition x.

Then 3^{2}

9 This also satisfies the condition x^{2}.

Hence Q(3,8) exists and it is true.

d)

Assume Q(x,y)=Q(9,10)

Here 9 satisfies the condition x

Then 9^{2}

81 satisfies the condition x^{2}.

Thus, Q(9,10) point exists and it is true. This satisfies the same values as in part (c)

6 0
3 years ago
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