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timofeeve [1]
3 years ago
10

Tyler wrote checks on his checking account for $20, $20, $18, $20, and $35. He also deposited $53 in the account. Which number d

escribes the overall change in the balance of his account? Your answer: −$113 $60 −$60
Mathematics
1 answer:
bagirrra123 [75]3 years ago
6 0
$60 I think
Use a calculator just in case I’m wrong
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Pleas help me please.
Nezavi [6.7K]

Answer:

Step-by-step explanation:

When the circle is fill in it means theres more than one answer so it can be greater than or equal to or less than or equal to. When theres only a circle is means greater then or less than

I hope this helps you to get the right answers.

5 0
3 years ago
Is every relation also a function? Explain
Sladkaya [172]

Answer:

no

Step-by-step explanation:

they are not

8 0
2 years ago
Read 2 more answers
There are five identical blue books, two identical green books, and three identical black books. How many different patterns can
dsp73

Answer:

2520 patterns

Step-by-step explanation:

In 'n' 10!  ways, books can be arranged. But, there are also 5! permutation of blue books 'n1', 2! permutation of identical green books 'n2', and 3! permutation identical black books 'n3'.

Therefore, for non identical arrangements:

\frac{n!}{n1!n2!n3!}

\frac{10!}{5!2!3!} = 2520

Therefore, the books can be arranged on a shelf in 2520 patterns

8 0
3 years ago
Which of the following is the correct factorization of the polynomial below? x^3-15
Anuta_ua [19.1K]
The polynomial is irreducible.
4 0
3 years ago
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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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