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wariber [46]
3 years ago
6

PLZ HURRY ITS TIMEDDDDD

Mathematics
2 answers:
GaryK [48]3 years ago
8 0

A. 1/5k - 2/3j and -2/3j + 1/5k

Step-by-step explanation:

Equivalent expressions are expressions that are the same even though they may appear different.When you plug in a value to represent a variable in these expression, they will give same answer when simplified.

In

1/5k - 2/3j and -2/3j + 1/5k you notice the operation signs in the terms has been maintained though the position of the terms shifted.

Learn More

Equivalent expressions :brainly.com/question/280220

Keywords: expressions, equivalent

#LearnwithBrainly

swat323 years ago
8 0

Answer:

The answer is A!

Happy to help! :}

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Brian invests £1500 into his bank account. He receives 2% per year simple interest. How much will Brian have after 3 years? Give
Jet001 [13]

Answer:

1590

Step-by-step explanation:

1500 x 0.02

= 30

30 x 3

= 90

1500+90 = 1590

(ps.. there was an equation for this i just forgot it, you can look it up and check it)

8 0
3 years ago
Clearing fractions<br><br> SOLVE
lana66690 [7]
You can multiply by -30 to clear fractions.
.. 47 ≤ 10x +60

Now, subtract 60 and divide by 10
.. -13 ≤ 10x
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3 years ago
There is a bag filled with 2 blue, 4 red and 3 green marbles.
NISA [10]

Answer:

5 out of 9

Step-by-step explanation:

4 0
2 years ago
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A
KATRIN_1 [288]

Answer:

D. Zacpszcap. is probably the answer

7 0
2 years ago
PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST
ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

7 0
3 years ago
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