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tekilochka [14]
3 years ago
13

What is C/a + 4 when a=2, and c= 10

Mathematics
2 answers:
tigry1 [53]3 years ago
6 0

Answer:

4 and 10/2 so actually 9

Step-by-step explanation:

kumpel [21]3 years ago
4 0

Answer:

The answer is 9.

Step-by-step explanation:

You just need to substitute the variables, so:

10/2+4

= 9

I hope this helps!

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Ted needs an average of at least 70 on his three history tests. He has already scored 85 and 60 on two tests. What is the minimu
baherus [9]
He must get a 65 as a minimum grade on his third test to get an average of 70.
Equation is
                       85 + 60 + x
                     ------------------- greater than or equal to 70
                              3
3 0
3 years ago
What is the recursive formula of the arithmetic sequence 0,11,22,33,
ArbitrLikvidat [17]

Answer:

f(1)=0

f(x+1)=f(x)+11

7 0
3 years ago
11 5/8 +9 1/2 is my problem
Leto [7]

Answer:21 1/8 is the answer

Step-by-step explanation:

6 0
2 years ago
Which of the x-values are solutions to the following inequality? x< 100 Help ASAP
netineya [11]

Answer:

x = - infinity to 99

Step-by-step explanation:

The values of x must less than 100.

3 0
3 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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