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Mama L [17]
2 years ago
7

Triangle L M Q is cut by perpendicular bisector L N. Angle N L Q is 32 degrees and angle L M N is 58 degrees.

Mathematics
1 answer:
Serga [27]2 years ago
8 0

Answer:

Yes

Step-by-step explanation:

ΔMNL ≅ ΔQNL  by ASA or AAS

by ASA

Proof:

∠ LNM = ∠LNQ    =90

LN = LN   {Common}

∠MLN = ∠QLN     {LN bisects ∠ L}

By AAS

∠Q + ∠QLN + ∠LNQ = 180  {Angle sum property of triangle}

∠Q + 32 + 90 = 180

∠Q  + 122 = 180

∠Q = 180 -122 =

∠Q = 58

∠Q = ∠M

∠MNL =∠QNL = 90

LN = LN {common side}

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Brandon bought 21 chicken wings for $39.90. How much would it cost for 15 wings?
Pachacha [2.7K]

Answer:

$28.50

Step-by-step explanation:

21 chicken wings= $39.90

1 chicken wing= 39.90÷21= $1.90

15 chicken wings= 15×1.90= $28.50

4 0
2 years ago
Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to
DiKsa [7]

Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:  

p_v =P(Z  

c) If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

Step-by-step explanation:

Data given and notation  

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

\hat p=0.35 estimated proportion of business owners plan to provide a holiday gift to their employees

p_o=0.46 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:  

Null hypothesis:p\geq 0.46  

Alternative hypothesis:p < 0.46  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

Part c

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

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Check the attached file for the solution.

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Answer:

Step-by-step explanation:

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Its effect is such that the mean becomes extremely sensitive to extreme outliers making it possible that the mean is this not a representative of the population and this theoretically affects the standard deviation.

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Answer:

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