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2 years ago

Need step by step help

Mathematics

2 answers:

Llana [10]2 years ago

6 0

Answer:

E = 53°

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<em>good luck, i hope this helps :)</em>

MrRa [10]2 years ago

4 0

Answer:

what do u think it is?

Step-by-step explanation:

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What is the value of y?

mrs_skeptik [129]

Ok so you would start by getting rid of the fraction. You should know the steps for that by now but here this is what it would look like
3y/4+10=2y/4 Simplify
3y/4+10=1y/2
10=-1y
I really simplified the last part but this is what it looks like: $\frac{2y}{4}$ - $\frac{3y}{4}$ = -1y
10=-1y
y=-10

8 0

2 years ago

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How do you find measure a and c

astra-53 [7]

You can use the measure of 112 on the similar shape to find b, as they are same side exterior angles. From there, you can find both a and c, because a/b and b/c are supplementary angles.

a = 68 degrees

c = 68 degrees

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2 years ago

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jake said he has 8/100 of a dollar in his pocket Margie said that the decimal for 8/100 is 0.8 what error did she make. what is

lukranit [14]

The 8 is in the tenths place it would look like 8/10 .8 its .08 for 8/100

4 0

2 years ago

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Assume a warehouse operates 24 hours a day. Truck arrivals follow Poisson distribution with a mean rate of 36 per day and servic

kirill [66]

The expected waiting time in system for typical truck is 2 hours.

Step-by-step explanation:

Data Given are as follows.

Truck arrival rate is given by, α = 36 / day

Truck operation departure rate is given, β= 48 / day

A constructed queuing model is such that so that queue lengths and waiting time can be predicted.

In queuing theory, we have to achieve economic balance between number of customers arriving into system and that of leaving the system whether referring to people or things, in correlating such variables as how customers arrive, how service meets their requirements, average service time and extent of variations, and idle time.

This problem is solved by using concept of Single Channel Arrival with exponential service infinite populate model.

Waiting time in system is given by,

$w_{s} = \frac{1}{\alpha - \beta }$

where $w_s$ is waiting time in system

$\alpha$ is arrival rate described Poission distribution

$\beta$ is service rate described by Exponential distribution

$w_{s} = \frac{1}{\alpha - \beta }$

$w_{s} = \frac{1}{48 - 36 }$

$w_{s} = \frac{1}{12 } day$

$w_{s} = \frac{1}{12 } \times 24 hour$ ...it is due to 1 day = 24 hours

$w_{s} = 2 hours$

Therefore, time required for waiting in system is 2 hours.

5 0

3 years ago

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

4 0

2 years ago