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Over [174]
3 years ago
11

What does it mean for rows to be linearly independent?

Mathematics
1 answer:
garri49 [273]3 years ago
4 0
It means no rows in the set can be written or defined as a linear combination of the others.
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What is the value of x, given that figure LMNO is a square? A. 45 B. 33 C. 27 D. 81
Degger [83]
The is answer is B. 33
8 0
3 years ago
Find x .<br><br> 8<br> 4√2<br> √8<br> 16
love history [14]

Answer:

8

Step-by-step explanation:

(8\sqrt{2}) ^{2} = 128 \\128 / 2 = 64\\x= \sqrt{64} = 8

8 0
2 years ago
Please help with this on the picture
galina1969 [7]
Answer:

a) 81

b) 16,807

c) 100

d) 78125

Step-by-step explanation:

a)

3^2 x 3^2

Step 1. Simplify the exponents

3^2 = 3 x 3 = 9

Step 2. Multiply

9 x 9 = 81
••••••••••••••••••••••••••••••••

b)

7 x 7^4

Step 1. Simplify the exponent

7^4 = 7 x 7 x 7 x 7 = 2401

Step 2. Multiply

7 x 2401 = 16,807
••••••••••••••••••••••••••••••••

c)

10^9 divided by 10^7

Step 1. Simplify the exponents

10^9 = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1000000000

10^7 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10000000

Step 2. Divide

1000000000 divided by 10000000 = 100
••••••••••••••••••••••••••••••••

d)

5^8 divided by 5

Step 1. Simplify the exponent

5^8 = 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 = 390625

Step 2. Divide

390625 divided by 5 = 78125



7 0
2 years ago
What is the measurement of the unknown sixth interior angle of this hexagon?
uranmaximum [27]
Can I see a photo can’t tell what you are talking about
7 0
3 years ago
Read 2 more answers
A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random
Alenkasestr [34]

Answer:

a. Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

b. t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

c. p_v =P(t_{97}>2.287) =0.0122

So with 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% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 >\mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 \leq 0

Alternative hypothesis: \mu_1 -\mu_2>0

Our notation on this case :

n_1 =78 represent the sample size for group 1

n_2 =21 represent the sample size for group 2

\bar X_1 =92.88 represent the sample mean for the group 1

\bar X_2 =86.90 represent the sample mean for the group 2

s_1=15.34 represent the sample standard deviation for group 1

s_2=8.99 represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom. If we replace the values given we have:

t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282

Now we can calculate the degrees of freedom given by:

df=78+21-2=97

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

p_v =P(t_{97}>2.287) =0.0122

So with 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% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).  

6 0
3 years ago
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