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GaryK [48]
3 years ago
9

PLEASE HELP

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
8 0
x= -9, 9


you have to subtract 15
then u will get x^2= 81
square root on both sides

And u get 9
Katarina [22]3 years ago
5 0

Answer:

Step-by-step explanation:

15 + c² = 96

c²  = 96 - 15

c²  = 81

c²  = 9²

c = 9

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Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minim
andreev551 [17]

Answer:

p_v =P(t_{74}    

If we compare the p value and a significance level for example \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.

Step-by-step explanation:

Data given and notation    

\bar X=18.81 represent the average lateral recumbency for the sample    

s=8.4 represent the sample standard deviation    

n=75 sample size    

\mu_o =20 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

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

State the null and alternative hypotheses.    

We need to apply a left tailed  test.  

What are H0 and Ha for this study?    

Null hypothesis:  \mu \geq 20  

Alternative hypothesis :\mu < 20  

Compute the test statistic  

The statistic for this case is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

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

t=\frac{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227

The degrees of freedom are given by:

df=n-1=75-1=74    

Give the appropriate conclusion for the test  

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

p_v =P(t_{74}    

Conclusion    

If we compare the p value and a significance level for example \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.

3 0
3 years ago
Dannette and Alphonso work for a computer repair company. They must include the time it takes to complete each repair in their r
olga55 [171]

Answer:

(a):

Dannette                   Alphonso

\bar x_D = 4.33                    \bar x_A = 5.17

M_D = 2.5                    M_A = 5

\sigma_D = 3.350                  \sigma_A = 1.951

IQR_D = 7                  IQR_A = 1.5

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

A = \{3,4,4,4,4,5,5,5,5,6,6,11\} ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

\bar x = \frac{\sum x}{n}

So, we have:

\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}

\bar x_D = \frac{52}{12}

\bar x_D = 4.33 --- Dannette

\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}

\bar x_A = \frac{62}{12}

\bar x_A = 5.17  --- Alphonso

Median

The median is calculated as:

M = \frac{n + 1}{2}th

M = \frac{12 + 1}{2}th

M = \frac{13}{2}th

M = 6.5th

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

M_D = \frac{2+3}{2}

M_D = \frac{5}{2}

M_D = 2.5 ---- Dannette

M_A = \frac{5+5}{2}

M_A = \frac{10}{2}

M_A = 5  ---- Alphonso

Standard Deviation

This is calculated as:

\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}

So, we have:

\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}

\sigma_D = \sqrt{\frac{134.6668}{12}}

\sigma_D = 3.350 ---- Dannette

\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}

\sigma_A = \sqrt{\frac{45.6668}{12}}

\sigma_A = 1.951 --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

IQR =Q_3 - Q_1

Where

Q_3 \to Upper Quartile       and        Q_1 \to Lower Quartile

Q_3 is calculated as:

Q_3 = \frac{3}{4}*({n + 1})th

Q_3 = \frac{3}{4}*(12 + 1})th

Q_3 = \frac{3}{4}*13th

Q_3 = 9.75th

This means that Q_3 is the mean of the 9th and 7th item. So, we have:

Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16           Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11

Q_3 = 8 ---- Dannette                 Q_3 = 5.5 --- Alphonso

Q_1 is calculated as:

Q_1 = \frac{1}{4}*({n + 1})th

Q_1 = \frac{1}{4}*({12 + 1})th

Q_1 = \frac{1}{4}*13th

Q_1 = 3.25th

This means that Q_1 is the mean of the 3rd and 4th item. So, we have:

Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2                  Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8

Q_1 = 1 --- Dannette                   Q_1 = 4 ---- Alphonso

So, the IQR is:

IQR = Q_3 - Q_1

IQR_D = 8 - 1                                     IQR_A = 5.5 - 4

IQR_D = 7 --- Dannette                      IQR_A = 1.5 --- Alphonso

Solving (b): The measures to compare

Measure of  center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of  spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

Lower =Q_1 - 1.5 * IQR

Upper =Q_3 + 1.5 * IQR

For Dannette:

Lower = 1 - 1.5 * 7 = -9.5

Upper = 8 + 1.5 * 7 = 18.5

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

Valid = 0 \to 18.5

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

Lower = 4 - 1.5 * 1.5 =1.75

Upper = 5.5 + 1.5 * 1.5 =7.75  

So, the valid data range are:

Valid = 1.75\to 7.75

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

4 0
3 years ago
Abraham went hiking last Saturday. From the car park, he walked 4 km due north,
jonny [76]

Answer:

a. (see attachment)

b. 6.5 km

c. 4.717 km

MAKE SURE TO USE UNITS

4 0
2 years ago
Please i need help i need 1-12
ANEK [815]
I gotchu. This is pretty simple okay? So when the problem is asking for X and X is inside the triangle, all you have to do is add the other two numbers, which are shown, and subtract it by 180 to get X. For example for problem number 1. You have x that’s inside the triangle, so you add up 47 and 58 which equals to 105 and then take that number and subtract it FROM 180. So the whole equation would be 47+58=105 180-105=75 so X=75. Now for other problems it’s the exact same thing all you have to do is take the other numbers add them up subtract it from 180 and you get X
5 0
2 years ago
Which model can represent the number 0.111
Dmitriy789 [7]
111/1000 or 11.1%. Any questions other than that?

7 0
2 years ago
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