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Dafna11 [192]
4 years ago
15

What is the value between 5 and 6

Mathematics
1 answer:
ololo11 [35]4 years ago
5 0
11 because if you add but if you multiple it is 30 because count by your 5's 5 10 15 20 25 30 see
You might be interested in
A student solves the following equation and determines that the solution is −2. Is the student correct? Explain. 3 a + 2 − 6a a2
Ganezh [65]

For this case, we have the following equation (according to the comments):

\frac{3}{a+2} - \frac{6a}{a^2-4}  = \frac{1}{a-2}

First, we factor the quadratic expression in the denominator:

\frac{3}{a+2} - \frac{6a}{(a-2)(a+2)}  = \frac{1}{a-2}

Then, we multiply both sides of the equation by (a-2) (a + 2):

\frac{3(a-2)(a+2)}{a+2} - \frac{6a(a-2)(a+2)}{(a-2)(a+2)}  = \frac{(a-2)(a+2)}{a-2}

Later, canceling similar terms we have:

3(a-2) - 6a = a+2

We do distributive property on the left side of the equation:

3a-6-6a=a+2

By grouping variables and constant terms we have:

3a - 6a -a = 2+6

Rewriting we have:

-4a=8

Finally, by clearing "a" we have:

a=-\frac{8}{4} a=-2

Note: the value of a is a extraneous solution because it makes the denominator of the original equation equal to zero.

Answer:

the student correct, but the value of a= -2 is an extraneous solution

8 0
4 years ago
Read 2 more answers
Madyson goes to a restaurant with her best friend, Taylor. The two girls order a meal, Madyson's meal costs $23.75 and Taylor's
7nadin3 [17]

Answer:30.75

$23.75 +$15.00=30.75

5+3=6

21=3

5 0
3 years ago
In ΔHIJ, h = 40 cm, ∠J=20° and ∠H=93°. Find the length of j, to the nearest centimeter.
DaniilM [7]

Given:

In ΔHIJ, h = 40 cm, ∠J=20° and ∠H=93°.

To find:

The length of j, to the nearest centimeter.

Solution:

According to Law of sine,

\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}

In ΔHIJ, using law of sine, we get

\dfrac{j}{\sin J}=\dfrac{h}{\sin H}

\dfrac{j}{\sin (20^\circ)}=\dfrac{40}{\sin (93^\circ)}}

j=\dfrac{40\times \sin (20^\circ)}{\sin (93^\circ)}}

On further simplification, we get

j=\dfrac{40\times 0.34202}{0.99863}

j=\dfrac{13.6808}{0.99863}

j=13.69958

Approximate the value to the nearest centimeter.

j\approx 14

Therefore, the length of j is 14 cm.

7 0
3 years ago
Rothamsted Experimental Station (England) has studied wheat production since 1852. Each year many small plots of equal size but
podryga [215]

Answer:

(a) 0.653

(b) 0.0198

(c) Yes, after testing we conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Step-by-step explanation:

(a) We are give the sample of one year annual production of wheat (in pounds) ;

4.46, 4.21, 4.40, 4.81, 2.81, 2.90, 4.93, 3.54, 4.16, 4.48, 3.26, 4.74, 4.97, 4.02, 4.91, 2.59

For calculating sample variance, firstly we will calculate mean of the above data;

Mean of above data, X_1bar = Sum of all values ÷ n (no. of values)

                                 = \frac{4.46 + 4.21+4.40+.......+4.91+2.59}{16} = 4.074

Sample Variance, s_1^{2} = \frac{\sum (X-X_1bar)^{2} }{n-1} = \frac{(4.46-4.074)^{2}+(4.21-4.074)^{2}+.........+(4.91-4.074)^{2}+(2.59-4.074)^{2} }{16-1} = 0.653

(b) Another sample for annual wheat production (in pounds);

3.89, 3.81, 3.95, 4.07, 4.01, 3.73, 4.02, 3.78, 3.72, 3.96, 3.62, 3.76, 4.02, 3.73, 3.94, 4.03

Mean of above data, X_2bar = Sum of all values ÷ n (no. of values)

                                 = \frac{3.89+3.81+3.95.......+3.94+4.03}{16} = 3.88

Sample Variance, s_2^{2} = \frac{\sum (X-X_2bar)^{2} }{n-1} = \frac{(3.89-3.88)^{2}+(3.81-3.88)^{2}+.........+(3.94-3.88)^{2}+(4.03-3.88)^{2} }{16-1} = 0.0198

(c) Now, we have to test the claim that population variance of annual wheat production for the first plot is larger than that for the second plot i.e.;

       Null Hypothesis, H_0 : \sigma_1^{2} = \sigma_2^{2}   or H_0 : \frac{\sigma_1^{2} }{\sigma_2^{2} } = 1

Alternate Hypothesis, H_0 : \sigma_1^{2} > \sigma_2^{2}   or H_0 : \frac{\sigma_1^{2} }{\sigma_2^{2} } >  1

The test statistics used here is;

                           \frac{s_1^{2} }{s_2^{2} }* \frac{\sigma_1^{2} }{\sigma_2^{2}}  ~ F_n__1-1_,n_2-1   where, n_1 = 16 and n_2 =16

   Test Statistics = \frac{0.653 }{ 0.0198}* 1 ~ F_1_5_,_1_5

                          = 32.98

Since, we are not provided with any significance level so we assume it to be 5% and at this level, the F table gives critical value of 2.4282.

<em>Since our test statistics is higher than the critical value and it falls in the rejection region so we have sufficient evidence to reject null hypothesis and conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.</em>

8 0
3 years ago
What's the answer and how do you know?!!
S_A_V [24]
C because its going back every time
7 0
4 years ago
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