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o-na [289]
3 years ago
9

Two rays with a common endpoint

Mathematics
2 answers:
nexus9112 [7]3 years ago
8 0

The common endpoint of two rays at which an angle is formed. Vertical Angles - Pairs of angles formed where two lines intersect. These angles are formed by rays pointing in opposite directions, and they are congruent.

Slav-nsk [51]3 years ago
6 0

This is an angle.  

start at a point and the rays shoot off.

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A company that makes​ hair-care products had 8,000 people try a new shampoo. Of the 8,000 ​people, 32 had a mild allergic reacti
Yuri [45]

Answer:

i think it 0.4%

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Construct a 98% confidence interval to determine whether there is any difference between the mean drug concentration in tablets
inna [77]

Answer:

Step-by-step explanation:

So, we are given two sites;

Site one: 91.28 , 92.83, 89.35, 91.90 82.85, 94.83, 89.83, 89.00, 84.62, 86.96, 88.32, 91.17, 83.86, 89.74, 92.24, 92.59, 84.21, 89.36, 90.96, 92.85, 89.39, 89.82, 89.91, 92.16, 88.67.

=> The mean = add up all the numbers/ the total numbers (which is 25). = 89.55.

=> The standard deviation = 3.005

SITE TWO: 89.35 86.51 89.04 91.82 93.02 88.32 88.76 89.26 90.36 87.16 91.74 86.12 92.10 83.33 87.61 88.20 92.78 86.35 93.84 91.20 93.44 86.77 83.77 93.19 81.79.

=> The mean= add up all the numbers/ the total numbers (which is 25). = 89.033.

=> Standard deviation = 3.271.

NB: the assumption here is that the variance for both sites are the same.

STEP ONE: DETERMINE THE DEGREE OF FREEDOM AND THE CRITICAL VALUE.

The degree of freedom =( mean of site 1 ) + (mean of site 2) - 2 = (25 + 25) − 2 = 48.

Since the degree of freedom is 48, Recall that the value for alpha given = 0.02. Then, the critical value = 2.407.

STEP TWO: DETERMINE THE STANDARD DEVIATION AND THE STANDARD ERROR FOR BOTH.

Therefore, the standard deviation = √[( 25 - 1) × 3.005^2 + (25 -1) × 3.271^2) ÷ (25 +25 - 2) ] = 3.14.

Also, the standard error = standard deviation × ( √(1/ mean 1 + 1/mean 2) ..

Standard error = 3.14 × [ ✓( 1/25 + 1/25)] = 0.89.

STEP THREE: DETERMINE THE CONFIDENCE INTERVAL.

confidence interval = (89.548 - 89.033 - 2.41 × 0.89, 89.548 - 89.033 + 2.41 × 0.89) = (- 1.62, 2.65).

STEP FOUR: THE HYPOTHESES.

=> THE TWO TAILED TEST.

Jo:μ1​ = μ2​.

Ja: μ1​ ≠ μ2​.

Thus, the rejected region is {t : ∣ t ∣ > 2.7}

As critical value = 2.7 for the degree of freedom which is 48 and α=0.01.

If we do the t - test for this,.we will have;

(89.548 - 89.003)÷ ( standard deviation × √ ( 1/25 + 1/25). = 0.58.

So,from our t - value, is less or equals to the critical value, therefore, the null hypothesis is not rejected.

Also, from the p-value table which gives p=0.565. The p- value is greater than or equal to the alpha value(0.01). therefore, the null hypothesis is not rejected.

We are not going to reject the null hypothesis because the information gathered is not enough.

3 0
2 years ago
Solve each of the following equations. (2 answers)<br><br> |2x-3|=2
Roman55 [17]

Answer:

x= 5/2 or 1/2

Step-by-step explanation:

absolute value always gives us positive

if |u| = 2, u can be -2 or 2

2x-3 = 2

2x = 5

x = 5/2

2x-3 = -2

2x = 1

x= 1/2

in general, set the stuff inside the absolute value to equal both positive and negative of the equivalent value

6 0
3 years ago
1. cot x sec4x = cot x + 2 tan x + tan3x
Mars2501 [29]
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
    cot(x)sec⁴(x)            cot(x)sec⁴(x)
                   0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
                   0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
                   0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
                   0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
                   0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
                   0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
                   0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
                   0 = cos⁴(x)(1 + tan²(x))²
                   0 = cos⁴(x)        or         0 = (1 + tan²(x))²
                ⁴√0 = ⁴√cos⁴(x)      or      √0 = (√1 + tan²(x))²
                   0 = cos(x)         or         0 = 1 + tan²(x)
         cos⁻¹(0) = cos⁻¹(cos(x))    or   -1 = tan²(x)
                 90 = x           or            √-1 = √tan²(x)
                                                         i = tan(x)
                                                      (No Solution)

2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
              sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
   sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
                               sin²(x) - cos²(x) = sin²(x) - cos²(x)
                                         + cos²(x)              + cos²(x)
                                             sin²(x) = sin²(x)
                                           - sin²(x)  - sin²(x)
                                                     0 = 0

3. 1 + sec²(x)sin²(x) = sec²(x)
           sec²(x)             sec²(x)
      cos²(x) + sin²(x) = 1
                    cos²(x) = 1 - sin²(x)
                  √cos²(x) = √(1 - sin²(x))
                     cos(x) = √(1 - sin²(x))
               cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
                                 x = 0

4. -tan²(x) + sec²(x) = 1
               -1               -1
      tan²(x) - sec²(x) = -1
                    tan²(x) = -1 + sec²
                  √tan²(x) = √(-1 + sec²(x))
                     tan(x) = √(-1 + sec²(x))
            tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
                             x = 0
5 0
3 years ago
3 connecting lines are shown. Line D F is horizontal. Line D E is about half the length of line D F. Line F E is about one-third
mars1129 [50]

The inequality that explains why the three segments cannot be used to construct a triangle is ED + EF < DF

<h3>Inequalities </h3>

From the question, we are to determine which of the given inequalities explains why the three segments cannot be used to construct a triangle

From the given information,

Line DE is about half the length of line DF

That is,

ED = 1/2 DF

Also,

Line FE is about one-third of the length of line DF

That is,

EF = 1/3 DF

Then, we can write that

ED + EF = 1/2DF + 1/3DF

ED + EF = 5/6 DF

Since,

5/6 DF < DF

Then,

ED + EF < DF

Hence, the inequality that explains why the three segments cannot be used to construct a triangle is ED + EF < DF

Learn more on Inequalities here: brainly.com/question/1447311

#SPJ1

5 0
2 years ago
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