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Firlakuza [10]
4 years ago
7

\begin{aligned} &-5y=-5 \\\\ &7x+6y=7 \end{aligned}

Mathematics
1 answer:
SSSSS [86.1K]4 years ago
7 0

Answer:

(5,1) is not a solution of the above system of equations.

Step-by-step explanation:

We are given a system of equations that are  

- 5y = - 5 ......... (1) and  

7x + 6y = 7 ........... (2)

So, solving equation (1) we get y = 1 and putting y = 1 in the equation (2) we get 7x = 7 - 6(1) = 1

⇒ x = \frac{1}{7}

So, it clear from the solution of the above equations i.e. (\frac{1}{7}, 1), we can say that (5,1) is not a solution of the above system of equations. (Answer)

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Using the following image, if E is the midpoint of FD find ED.<br><br>What is ED?
ELEN [110]

Answer:

10-5=5

Step-by-step explanation:

FD-FE=ED

FD=10

FE=5

10-5=5

ED=5

3 0
2 years ago
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In 2 hours golden completed 3\4 of his race. How long will it take Holden to complete the entire race
MrRa [10]
He would finish around 2 hrs and 40 minutes
4 0
4 years ago
PLZ HELP ILL GIVE 15 POINTS!!
Anna35 [415]

Answer:

There would be 14 cats and 9 dogs

Step-by-step explanation:

In order to find the amount of cats and dogs, we need to set up the system of equations. To do so, start by setting cats as x and dogs as y. Now we can write the first equation to show the total number of animals.

cats + dogs = 23

x + y = 23

Now we can write a second one that shows the difference in the number of cats and dogs.

cats - dogs = 5

x - y = 5

Now we can add the two equations together to solve for x.

x + y = 23

x - y = 5

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x = 14

Now that we have the number of cats, we can find the number of dogs by using either equation.

x + y = 23

14 + y = 23

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7 0
4 years ago
The amount of time it takes a bat to eat a frog was recorded for each bat in a random sample of 12 bats. The resulting sample me
spin [16.1K]

Answer: a. CI for the mean: 17.327 < μ < 26.473

b. CI for variance: 29.7532 ≤ \sigma^{2} ≤ 170.9093

Step-by-step explanation:

a. To construct a 95% confidence interval for the mean:

The given data are:

mean = 21.9

s = 7.7

n = 12

df = 12 - 1 = 11

1 - α = 0.05

\frac{\alpha}{2} = 0.025

t-score = t_{0.025,11} = 2.2001

Note: since the sample population is less than 30, it is used a t-score.

The formula for interval:

mean ± t.\frac{s}{\sqrt{n} }

Substituing values:

21.9 ± 2.200.\frac{7.7}{\sqrt{12} }

21.9 ± 4.573

The interval is: 17.327 < μ < 26.473

b. A 95% confidence interval for the variance:

The given values are:

s^{2} = 7.7^{2}

s^{2} = 59.29

α = 0.05

\frac{\alpha}{2} = 0.025

1-\frac{\alpha}{2} = 0.975

\chi^{2}_{0.025,11} = 21.92

\chi^{2}_{0.975,11} = 3.816

Note: To find the values for \chi^{2}_{\alpha/2,n-1} and \chi^{2}_{1-\alpha/2,n-1}, look for them at the chi-square table

The formula to calculate interval:

(\frac{(n-1).s^{2}}{\chi^{2}_{\alpha/2,n-1}} , \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2,n-1}})

are the lower and upper limits, respectively.

Substituing values:

(\frac{11.59.29}{21.92} , \frac{11.59.29}{3.816})

(29.7532, 170.9093)

The interval for variance is: 29.7532 ≤ \sigma^{2} ≤ 170.9093

6 0
3 years ago
Am I correct? The bottom answer is 4 if you can’t see it.
ivann1987 [24]

Answer:

yes i believe you didnt mess it up. goodluck mate

Step-by-step explanation:

4 0
4 years ago
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