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3 years ago

1. Find the lengths of the diagonals of rectangle QRST , Given that QS = 6X+3 and RT = 8X-1

Mathematics

2 answers:

Bezzdna [24]3 years ago

8 0

For the lengths, Qs and RT= 15

Greeley [361]3 years ago

5 0

Answer:

1. QS and RT= 15

2. Im not positive but I think it's TQ=18

Step-by-step explanation:

For question 1, you would do 6x+3=8x-1 and solve to get the result of x=2

Then you'd plug in 6 and get QS and RT=15

For the second one I am unsure because I wasn't positive how to solve but it seems like the side RQ might be 1/2 of TQ. Sorry if thats wrong I can try and help more if you need

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4 years ago

100 POINTS PLEASE HELP ME I BEG 100 POINTS

Marrrta [24]

Answer:

see explanation

Step-by-step explanation:

(1)

Since FE = FG the triangle is isosceles with ∠ E = ∠ G, then

∠ E = $\frac{180-106}{2}$ = $\frac{74}{2}$ = 37°

(2)

Since all 3 sides are congruent then triangle is equilateral with the 3 angles being congruent, 60° each , then

12y = 60 ( divide both sides by 12 )

y = 5

(3)

The 3 angles are congruent then triangle is equilateral with the 3 sides being congruent, then

KL = JL , that is

4t - 8 = 2t + 1 ( subtract 2t from both sides )

2t - 8 = 1 ( add 8 to both sides )

2t = 9 ( divide both sides by 2 )

t = 4.5

(4)

Given ∠ B = ∠ C then triangle is isosceles with 2 legs being congruent , that is

AB = AC

4x + 1 = 9 ( subtract 1 from both sides )

4x = 8 ( divide both sides by 4 )

x = 2

Then

perimeter = AB + BC + AC = 4x + 1 + 2x + 3 + 9

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= 6(2) + 13

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7 0

2 years ago

Read 2 more answers

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05