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alexgriva [62]
2 years ago
13

Based on the graph, what is the initial value of the linear relationship?

Mathematics
1 answer:
Leya [2.2K]2 years ago
6 0

Answer:

please help me please yh the picture and answers

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What is the measure of m
Ede4ka [16]
117 degrees

The angles in a trapezoid add up to 360 degrees. 63 x 2 = 126

360 - 126 = 234

234/2 = 117
5 0
3 years ago
A new youth sports center is being built in a safe harbor. The perimeter of the rectangular playing field is 510 yards. The leng
podryga [215]

Answer:

<h2>L=203 yards</h2><h2>W=52 yards</h2>

Step-by-step explanation:

Step one:

given data

perimeter= 510 yards

let the width be x, width=x

length= (4x-5)-----quadruple mean 4 times

Step two:

the expression for perimeter is

P=2L+2W

510=2(4x-5)+2x

510=8x-10+2x

510+10=8x+2x

520=10x

divide both sides by 10

x=520/10

x=52

the width is 52 yards

the lenght is  (4x-5)

L=4(52)-5

L=208-5

L=203 yards

4 0
3 years ago
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?​
zimovet [89]

Answer:

0.45 cm

Step-by-step explanation:

Actual length of the micro chip

=  \frac{1}{40}  \times 18 \\  \\  =0.45 \: cm

4 0
3 years ago
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

       T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
PLEASE HELP PLEAsee before It’s due
xz_007 [3.2K]
The measure of BOC is 47
6 0
3 years ago
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