Which procedure would you use to convert a fraction to a decimal?1
Convert to a decimal.
5
to a decimal?
Answers
Hope this helps!
Related Questions
In some division problems, a number or pattern of number that continues indefinitely is a?
In the sixth grade class at Madison Middle School, for every 6 students, 4 of them play at least one sport. If there are 270 students on the sixth grade class, how many of them play at least one sport? a) 39b) 180c) 11d) 405
YThe figure shows A XYZ. XW is the anglebisector of ZYXZ.86.5WWhat is W Z?Enteryour answer in the box. Do not roundyour answer.xZ6unitsBasic
What is 9 + 2x = -x + 3
Answers
Answer:
y = -1.
Step-by-step explanation:
The value of y is -1 as it passes through the y axis. The values of x on the line are infinite.
This line's equation is y = -1.
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Answer:
12:15
Step-by-step explanation:
Answers
The number of federally insured banks in the year 1990 was 12,100.
Given data:
To find the number of federally insured banks in the year 1990, you need to substitute the value of t = 1990 into the given function B(t):
B(t) = -329.4t + 13747
Since t corresponds to the year and t = 0 corresponds to 1985, we need to calculate for t = 1990 - 1985.
t = 5.
Substitute t = 5 into the function:
B(5) = -329.4 * 5 + 13747
Now, perform the calculations:
B(5) = -1647 + 13747
B(5) = 12100
Hence, there were approximately 12,100 federally insured banks in the year 1990.
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Final answer:
Using the given equation and by substituting t with 5 (for the year 1990, five years after 1985) in that equation, we find that there were 13363 federally insured banks in 1990.
Explanation:
The question is asking how many federally insured banks there were in 1990, given that the number of banks in any given year from 1985 to 2007 is given by the equation
B(t) = -329.4t + 13747
. In this equation,
t
is the year with t=0 corresponding to 1985. To find the answer, we should substitute in the appropriate value of t, which would be 5 because 1990 is five years after 1985. Thus, the equation becomes B(5) = -329.4*5 + 13747. This simplifies to 15010 - 1647, which equals
13363
. Therefore, there were 13363 federally insured banks in 1990.
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Answers
Answer:
sorry i dont have insta
Step-by-step explanation:
hope you find help!
Answer:
What do u need help with
Answers
Answer:
229.50 ÷ 8.50= 27 people
b. Using a critical value, test the null hypothesis at the 5% level of significance.
c. Using a p-value, test the hypothesis at the 5% level of significance.
d. What type of error may have been committed for this hypothesis test
Answers
Answer:
a)
b) The z-value (1.5) is smaller than z=1.645 (critical value), so it is in the "acceptance region". It failed to reject the null hypothesis.
c) The p-value (0.07) is greater than the significance level (0.05), so it failed to reject the null hypothesis.
d) In this case, the error we may hae comitted is a Type II error (failed to reject a null hypothesis that is false).
Step-by-step explanation:
We have to perfomr a hypothesis test on the mean, with known standard deviation of the population.
a) The null hypothesis is that the deliver time is 15 days or less.
The null and alternative hypothesis are then:
The significance level is defined as 0.05.
b) The critical value of z for a one-side test (rigth side) and a significance level of 0.05 is z=1.645.
If the z-value for this sample is higher than 1.645, it is in the "rejection region".
Calculating the z-value:
The z-value (1.5) is smaller than z=1.645 (critical value), so it is in the "acceptance region". It failed to reject the null hypothesis.
c) The p-value for z=1.5 is:
The p-value (0.07) is greater than the significance level (0.05), so it failed to reject the null hypothesis.
d) There are two types of error:
Type I errors: happen when we reject a null hypothesis that is true.
Type II errors: happen when we failed to reject a null hypothesis that is false.
In this case, the error we may hae comitted is a Type II error.
The null hypothesis at 5% significance level is 1.50
Data;
- n = 49
- mean = 16.2
- standard deviation = 5.6
Null and Alternative Hypothesis
The alternative hypothesis are
H_o = mean = 15
standard deviation = 5.6 days
using z-test,
n is greater than or equal to 30
Assuming standard deviation is known
z-critical value is 1.96 at 95% confidence level.
Since z-critical value is greater than the test statistic, so we tail to subject H_o.
There's no evidence that mean delivery time is different from 15 days.
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