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Mazyrski [523]
3 years ago
7

What is the standard form of the line through the given point: (3,5), with slope = -3

Mathematics
1 answer:
Flura [38]3 years ago
5 0
The answer is 3x + y - 14 = 0
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The histogram shows information about how 550 people travel to work.
lubasha [3.4K]

Its 1 person i promise you thats really easy.

3 0
2 years ago
The mean work week for engineers in start-up companies is claimed to be about 63 hours with a standard deviation of 5 hours. Kar
Phantasy [73]

Answer:

a. Mean

b. μ = μ₀

c. 11.5%

d. Yes

Step-by-step explanation:

The given parameters are;

The mean work week for engineers, μ₀ = 63 hours

The standard deviation, σ = 5 hours

The number of engineers in Kara's sample, n = 10 engineers

The responses given by the 10 engineers are;

70; 45; 55; 60; 65; 55; 55; 60; 50; 55

a. The given information which the newly hired is hoping to find out that it is not is the <em>mean</em>

<em />

b. The null hypothesis which is that the company claim is correct, is therefore;

Null hypothesis, H₀; μ = μ₀ = 63 hours

c. Kara's mean, \overline x is found as follows;

\overline x = (70 + 45 + 55 + 60 + 65 + 55 + 55 + 60 + 50 + 55)/10 = 57

\overline x = 57 hours

The Z-score is therefore;

Z=\dfrac{\overline x-\mu }{\sigma }

Z = (57 - 63)/5 = -1.2

From the Z-table, we have;

The p-value for P(\overline x ≤ 57) =  P(Z ≤ -1.2) = 0.11507

The probability as a percentage given to one decimal place, that the mean would be as low as Kara's mean, P(\overline x ≤ 57) = 11.5%

d. Given that the percent percentage chance (11.5%) that the mean could be as low as hers (\overline x = 57 hours) is higher than 10%, she should accept the claim.

3 0
3 years ago
What is the GCF of 32 and 52
FinnZ [79.3K]
The GCF of 32 and 52 is 4. Hope this helps! :)
7 0
4 years ago
Read 2 more answers
An accounting firm is trying to decide between IT training conducted in-house and the use of the third party consultants. To get
Jlenok [28]

Answer:

t=\frac{500-490}{\sqrt{\frac{48^2}{180}+\frac{32^2}{210}}}}=2.379  

p_v =2*P(t_{388}>2.379)=0.0178

Comparing the p value with the significance level provided \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average of the two groups.  

Step-by-step explanation:

Assuming the following info

IT Training     N   mean Stdev

In-house       210 $490 $32

Consultants  180 $500 $48

1) Data given and notation

\bar X_{I}=490 represent the mean for the sample of In-house

\bar X_{C}=500 represent the mean for the sample Consutants

s_{I}=32 represent the population standard deviation for the sample In-house

s_{C}=48 represent the population standard deviation for the sample Consultants

n_{I}=210 sample size for the group In-house

n_{C}=180 sample size for the group Consultants

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis:\mu_{I}=\mu_{C}

Alternative hypothesis:\mu_{I} \neq \mu_{C}

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{C}-\bar X_{I}}{\sqrt{\frac{s^2_{C}}{n_{C}}+\frac{s^2_{I}}{n_{I}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

With the info given we can replace in formula (1) like this:

t=\frac{500-490}{\sqrt{\frac{48^2}{180}+\frac{32^2}{210}}}}=2.379  

4) Statistical decision

First we need to calculate the degrees of freedom given by:

df=n_{C}+n_{I}-2=210+180-2=388[/tex[The sample is large enough to assume that the distribution is also normal.Since is a bilateral test the p value would be:&#10;[tex]p_v =2*P(t_{388}>2.379)=0.0178

Comparing the p value with the significance level provided \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average of the two groups.  

6 0
4 years ago
Solve by the linear combination method.<br><br> 6x – 2y = 2<br> –3x + 4y = 5
kykrilka [37]
6x - 2y = 2
-3x + 4y = 5....multiply by 2
-----------------
6x - 2y = 2
-6x + 8y = 10 (result of multiplying by 2)
-----------------add
6y = 12
y = 12/6
y = 2

6x - 2y = 2
6x - 2(2) = 2
6x - 4 = 2
6x = 2 + 4
6x = 6
x = 6/6
x = 1

solution is (1,2)
8 0
3 years ago
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