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xxMikexx [17]
3 years ago
6

Select a point that is a solution to the system of inequalities

Mathematics
2 answers:
padilas [110]3 years ago
7 0
The answer to your question is C.
inessss [21]3 years ago
5 0

wrong, the answer is (2,6)

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Two triangles, Triangle 1 and triangle 2, are similar. Is it a Dilation?
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Answer:

You can't be sure without an image.

Step-by-step explanation:

If one moves left or right, that's a horizontal translation.

If one moves up or down, it's a vertical translation.

A dialation is when the image gets smaller or bigger.

8 0
2 years ago
What theorem can be used to prove that the two triangles are congruent?
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Pythagorean theorem a^2+b^2=c^2
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3 years ago
There are 68 calories in 1 serving of cereal. If a bowl holds 7 ​servings, how many calories are in the​ bowl?
maria [59]

Answer:

476

Step-by-step explanation:

68 \times 7 = 476

because there are 68calories per serving and there are 7 servings

6 0
3 years ago
For a Cinco de Mayo party, Alan is placing sets of chips and salsa's throughput the house. He has 20 bags of chips and 15 jars o
NemiM [27]
HCF. The answer is 5
7 0
2 years ago
Read 2 more answers
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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