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fredd [130]
2 years ago
13

5a - 2(4a + 1) + 6(4a - 1)​ show work pls

Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
6 0

Answer:

21<em>a</em> − 8

Step-by-step explanation:

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HELP PLZ AGAIN. Same material. It is still timed. Will give 100 pts and brainly. This is vital to my math grade. Time is on the
ruslelena [56]

Answer:

y=4x+47

I wanted to wait for the person who originally said the answer but it's been a while and I kinda want the points. Sorry

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3 years ago
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Write (8,3)m=-7 in point slope and slope intercept form
Lerok [7]

Answer:

The slope intercept form is y=-7x+59

Step-by-step explanation:

y-y=m(x-x1)

y-3=-7(x-8)

y-3=-7x+56

+3 +3

y=-7x+59

8 0
3 years ago
A equal t added to 258<br><br><br>write the sentence as an equal ​
Deffense [45]

Answer:

A=t+258

Step-by-step explanation:

6 0
3 years ago
The body temperatures of adults are normally distributed with a mean of 98.6degrees° F and a standard deviation of 0.60degrees°
Schach [20]

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 98.6, \sigma = 0.6, n = 36, s = \frac{0.6}{\sqrt{36}} = 0.1

If 36 adults are randomly​ selected, find the probability that their mean body temperature is greater than 98.4degrees° F.

This is 1 subtracted by the pvalue of Z when X = 98.4. So

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

By the Central Limit Theorem

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

Z = \frac{98.4 - 98.6}{0.1}

Z = -2

Z = -2 has a pvalue of 0.0228

1 - 0.0228 = 0.9772

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

6 0
3 years ago
Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert
adell [148]

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12} = 60.67

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}

Put value in formula of Standard Deviation,

s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}} = 40.75

Step-3: Then, we have to find the critical value by chi-square.

X_{1-\alpha/2}^{2}=3.82

X_{1-\alpha/2}^{2}=21.92

Then, find the confidence interval which is 95%.

\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9

\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9

5 0
3 years ago
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