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Nitella [24]
3 years ago
11

L || m If m∠1=125 and m∠7=50 then m∠5= 5 55 75 105

Mathematics
1 answer:
Fed [463]3 years ago
5 0
M < 2 = 180 - m < 1 = 180 - 125  = 55 degrees
m < 2  = m < 5 ( alternate angles) 
so m < 5 = m < 2 = 55 degrees answer
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An element with mass 800 grams decays by 8.2% per day. How much of the element is remaining after 15 days,
Serjik [45]

Given:

The initial mass of an element is 800 grams.

Decay rate = 8.2% per day

Number of days = 15

To find:

The remaining element after 15 days.

Solution:

The exponential decay model is

y=a(1-r)^t

Where, a is the initial value r is the rate of interest and t is time period.

Putting a=800,r=0.082,t=15 in the above formula, we get

y=800(1-0.082)^{15}

y=800(0.918)^{15}

y=221.68188

y\approx 221.7

Therefore, the mass of the remaining element is 221.7 grams.

7 0
3 years ago
Solve for x: x/25 &gt; 5
WITCHER [35]

Answer:

x>125

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Does it open up or down and is it shifted to the left 2 units or to the right 2 units of function g(x)
mixas84 [53]
It opens down if the lead coefficient is negative

And it shifts to the left by two. Always use the opposite of what is in parenthesis, which in this case is 2. 
7 0
3 years ago
A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po
igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b) Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c)z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation    

X_{1}=170 represent the number of people with the characteristic 1

X_{2}=110 represent the number of people with the characteristic 2  

n_{1}=200 sample 1 selected  

n_{2}=150 sample 2 selected  

p_{1}=\frac{170}{200}=0.85 represent the proportion estimated for the sample 1  

p_{2}=\frac{110}{150}=0.733 represent the proportion estimated for the sample 2  

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

p_v represent the value for the test (variable of interest)  

\alpha=0.05 significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

a.State the decision rule.

For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c. Compute the value of the test statistic.                                                                                              

z-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.    

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

5 0
3 years ago
The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c
Anestetic [448]

Answer:

a) The mean is \mu = 60

b) The standard deviation is \sigma = 9

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when X = 55.5, Z = -0.5

So

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

-0.5 = \frac{55.5 - \mu}{\sigma}

-0.5\sigma = 55.5 - \mu

\mu = 55.5 + 0.5\sigma

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when X = 71.52, Z = 1.28

So

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

1.28 = \frac{71.52 - \mu}{\sigma}

1.28\sigma = 71.52 - \mu

\mu = 71.52 - 1.28\sigma

Since we also have that \mu = 55.5 + 0.5\sigma

55.5 + 0.5\sigma = 71.52 - 1.28\sigma

1.78\sigma = 71.52 - 55.5

\sigma = \frac{(71.52 - 55.5)}{1.78}

\sigma = 9

\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60

Question

The mean is \mu = 60

The standard deviation is \sigma = 9

6 0
3 years ago
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