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2 years ago

Plz help with numbers 3 and 4 !!

Mathematics

2 answers:

vovangra [49]2 years ago

5 0

3.

30°

4. 100°. I think

goblinko [34]2 years ago

4 0

Answer:

3. opposite

30°

4. 50° + 30° + 2x = 180°

80° + 2x = 180°

2x = 100°

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Find m. DCR if m. RCB= 115°<br> and m_ DCB = 169º.

Burka [1]

It is 54. Draw it out. You will see that DCB - RCB = DCR

7 0

3 years ago

Allied Corporation is trying to sell its new machines to Ajax. Allied claims that the machine will pay for itself since the time

kvasek [131]

Answer:

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

Step-by-step explanation:

We must evaluate the differences of the means of the two machines, to do so, we will assume a CI of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).

New machine

Sample mean x₁ = 25

Sample variance s₁ = 27

Sample size n₁ = 45

Old machine

Sample mean x₂ = 23

Sample variance s₂ = 7,56

Sample size n₂ = 36

Test Hypothesis:

Null hypothesis H₀ x₂ - x₁ = d = 0

Alternative hypothesis Hₐ x₂ - x₁ < 0

CI = 90 % ⇒ α = 10 % α = 0,1 z(c) = - 1,28

To calculate z(s)

z(s) = ( x₂ - x₁ ) / √s₁² / n₁ + s₂² / n₂

s₁ = 27 ⇒ s₁² = 729

n₁ = 45 ⇒ s₁² / n₁ = 16,2

s₂ = 7,56 ⇒ s₂² = 57,15

n₂ = 36 ⇒ s₂² / n₂ = 1,5876

√s₁² / n₁ + s₂² / n₂ = √ 16,2 + 1.5876 = 4,2175

z(s) = (23 - 25 )/4,2175

z(s) = - 0,4742

Comparing z(s) and z(c)

|z(s)| < | z(c)|

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean

3 0

3 years ago

In a candy mix there are 4 green bars for every 3 red bars. How many red bars are there if there are 200 green bars

erastovalidia [21]

Answer:

Since there are 4 green bars for every 3 red bars and we are trying to find the number of red bars if there are 200 green bars, we can create the ratio:

4 x : 3 y

Where x is equal to the number of green bars and y is the number of red bars.

We know the number of green bars is equal to 200, so we can divide it by 4, giving us:

200 /4=50

Then we can solve for y , the number of red bars.

// Multiple y by 50

3 ⋅ 50 = 150

So for every 4 green bars, there are 3 red bars.

For every 200 green bars, there are 150 red bars

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees. Befor

Leya [2.2K]

Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 700, \sigma = 50$ .

The probability of a month having a PCE between $575 and $790 is the <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, hence:

X = 790:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{790 - 700}{50}$

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

X = 575:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{575 - 700}{50}$

Z = -2.5

Z = -2.5 has a p-value of 0.0062.

0.9641 - 0.0062 = 0.9579.

0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

More can be learned about the normal distribution at brainly.com/question/4079902

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3 0

1 year ago

What is the mode for the set of data? 6,7,10,12,12,13<br><br> 10<br> 12<br> 13<br> 11

KonstantinChe [14]

12
because it is the only number that appears more than once which is basically the definition of mode.

4 0

3 years ago