Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Response:

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<span>With AB tangent at D, where AD = OD = 4
, triangle OAD forms a right angle with sides measuring 4 and 4.
So, the area of triangle OAD is computed as 1/2 * 4 * 4 = 8

The angles AOD and DAO equal 45 degrees.
Therefore, the area of circular sector OCD is calculated as the area of circle O multiplied by 45/360
= pi * 4 * 4 * 45/360
= 2pi

The shaded area ACD results from the triangle OAD minus the area of circular sector OCD
= 8 - 2pi
= 1.72




</span>

We need to specify the interval for numbers ranging from -5 to 4.

Moreover, we must exclude the values -1 and 4 from this range.

First, let’s note some crucial guidelines regarding interval notation.

It’s essential to identify which values are included and which ones are not in the specified range.

An open interval (or) indicates excluded numbers, while a closed interval [or] shows included numbers.

In this scenario, we notice three key numbers: -5, -1, and 4.

It is explicit that -1 and 4 are excluded, though -5 is not indicated as excluded, hence it remains included.

-1 lies between -5 and 4 on the number line.

Consequently, we can represent the interval notation from -5 to -1. (Note: -5 is included and -1 is not).

Thus, we write a closed interval " [ " on the left side of -5 and an open interval " ( " on the right side of -1. This gives us,

[-5, -1 ).

Next, we represent the interval from -1 to 4.

Since both endpoints are excluded, we will use open intervals ( ) on both sides of -1 and 4.

This results in the interval being

(-1,4).

We combine the two interval notations using the union symbol. So, the final interval representation is

[-5, -1 ) ∪ (-1,4).

Answer:

We conclude that less than or equal to 50% of the student body supports him significantly.

Step-by-step explanation:

Mike, while campaigning for the student government presidency, is keen to determine if more than 50% of the student body supports him.

A random sample of 100 students was surveyed, with 55 expressing support for Mike.

Let p = the proportion of students backing Mike.

Thus, Null Hypothesis, : p 50% {indicating that the proportion of supporters among the student body is significantly less than or equal to 50%}

Alternate Hypothesis, : p > 50% {indicating that the proportion of supporters among the student body is significantly more than 50%}

The test statistics to be utilized here One-sample z proportion statistics;

T.S. = ~ N(0,1)

where, = the sampled proportion in favor of Mike = = 0.55

n = number of sampled students = 100

Therefore, test statistics =

= 1.01

The z test statistic is 1.01.

At a significance level of 0.05, the z table provides a critical value of 1.645 for a right-tailed test.

Given that our test statistic falls below the critical value of z (1.01 < 1.645), we lack sufficient evidence to reject the null hypothesis as it remains outside the rejection region, leading to failure to reject our null hypothesis.

As a result, we conclude that less than or equal to 50% of the student body is in favor of him or the proportion supporting Mike is not significantly greater than 50%.