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frozen [14]
2 years ago
15

Solve the following quadratic equations by extracting square roots.Answer the questions that follow.

Mathematics
2 answers:
Delicious77 [7]2 years ago
8 0

Answer:

1.+4,-4\\2. +9,-9\\3. +10,-10\\4. +12, -12\\5. +5, -5

Step-by-step explanation:

IN order to solve the quadratic equations you just have to solve the square root of the numeric part of the equation:

x^{2} =16\\x=\sqrt{16}\\ x= +4, -4

t^{2} =16\\t=\sqrt{81}\\ x= +9, -9

r^{2} =100\\r=\sqrt{100}\\ x= +10, -10

x^{2} -144=0\\x=\sqrt{144}\\ x= +12, -12

2s^{2}=50\\s^{2}=\frac{50}{2} \\s=\sqrt{25}\\ s= +5, -5

Just remember that the solution for any square root will always be a negative and a positive number.

Vikentia [17]2 years ago
4 0

Answer:

1.  x=±4

2. t=±9

3. r=±10

4. x=±12

5. s=±5

Step-by-step explanation:

1. x^2 = 16

Taking square root on both sides

\sqrt{x^2}=\sqrt{16}\\\sqrt{x^2}=\sqrt{(4)^2}\\

x=±4

2. t^2=81

Taking square root on both sides

\sqrt{t^2}=\sqrt{81}\\\sqrt{t^2}=\sqrt{(9)^2}

t=±9

3. r^2-100=0

r^{2}-100=0\\r^2 =100\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{r^2}=\sqrt{100}\\\sqrt{r^2}=\sqrt{(10)^2}

r=±10

4. x²-144=0

x²=144

Taking square root on both sides

\sqrt{x^2}=\sqrt{144}\\\sqrt{x^2}=\sqrt{(12)^2}

x=±12

5. 2s²=50

\frac{2s^2}{2} =\frac{50}{2}\\s^2=25\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{s^2}=\sqrt{25}\\\sqrt{s^2}=\sqrt{(5)^2}

s=±5 ..

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A quality control manager at an auto plant measures the paint thickness on newly painted cars. A certain part that they paint ha
Scorpion4ik [409]

Answer:

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 2, \sigma = 0.8, n = 100, s = \frac{0.8}{\sqrt{100}} = 0.08

What is the probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value?

This is the pvalue of Z when X = 2 + 0.1 = 2.1 subtracted by the pvalue of Z when X = 2 - 0.1 = 1.9. So

X = 2.1

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

By the Central Limit Theorem

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

Z = \frac{2.1 - 2}{0.08}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

X = 1.9

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

Z = \frac{1.9 - 2}{0.08}

Z = -1.25

Z = -1.25 has a pvalue of 0.1056

0.8944 - 0.1056 = 0.7888

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

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2 years ago
A polygon is graphed on a coordinate grid with (x,y) representing the location of a certain point on the polygon. the polygon is
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Step-by-step explanation: MARK ME BRAINLIEST!!!!

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2 years ago
Paul bought a package of 6 spiral notebooks for a total cost of $13.50. which equation represents p, the cost, in dollars, of ea
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The results of a mathematics placement exam at two different campuses of Mercy College follow: Campus Sample Size Sample Mean Po
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Answer:

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

Step-by-step explanation:

Data given

Campus   Sample size     Mean    Population deviation

   1                 330               33                      8

   2                310                31                       7

\bar X_{1}=33 represent the mean for sample 1  

\bar X_{2}=31 represent the mean for sample 2  

\sigma_{1}=8 represent the population standard deviation for 1  

\sigma_{2}=7 represent the population standard deviation for 2  

n_{1}=330 sample size for the group 1  

n_{2}=310 sample size for the group 2  

\alpha Significance level provided  

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

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.  

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

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

P value  

Since is a one right tailed test the p value would be:  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

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