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ruslelena [56]
3 years ago
15

Can you help me please?

Mathematics
2 answers:
daser333 [38]3 years ago
6 0

Answer:

125

Step-by-step explanation:

180-55

125

Reason: Co interior angles

Sergeeva-Olga [200]3 years ago
3 0

Answer:

ither 125 or none

Step-by-step explanation:

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h=2A/b

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PLEASE HELP IM DESPERATE WILL GIVE BRAINLIEST ONLY ANSWER IF YOU ARE HELPING
VARVARA [1.3K]

Answer:

48

Step-by-step explanation:

u • w

= (i + 6j ) • (9i - 2j )

= (1 × 9 ) + (6 × - 2 )

= 9 + (- 12)

= 9 - 12

= - 3

v • w

= (5i - 3j ) • (9i - 2j )

= (5 × 9) + (- 3 × - 2)

= 45 + 6

= 51

Then

u • w + v • w

= - 3 + 51

= 48

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3 years ago
Question 6 Financial Literacy Quiz
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Answer:

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Step-by-step explanation:

i just want the points

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Mario is cutting loaves of bread into slices each loaf of bread is 15 centimeters tall, 15 centimeters wide, and 30 centimeters
denis23 [38]
Divide the is 2. but not that sure
5 0
3 years ago
Researchers fed mice a specific amount of Dieldrin, a poisonous pesticide, and studied their nervous systems to find out why Die
Elodia [21]

Answer:

Step-by-step explanation:

Part A

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.

6 0
3 years ago
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