120,000 renamed to ten thousand (10,000) is twelve ten thousand. Ten thousand in number form is 10,000. When you divide 120,000 (one hundred and twenty thousand) by 10,000 (ten thousand) you get 12. 120,000 (one hundred and twenty thousand) /10,000 (ten thousand) = 12. So there are twelve ten thousands in 120,000.
Answer:
The loan was for 9 months only
Step-by-step explanation:
In this question, we are concerned with calculating the time taken for a loan om an interest to be paid back
To calculate this, we use the simple interest formula
Mathematically;
I = PRT/100
where P is the principal which is the amount borrowed and that is $500 according to the question
R is the rate which is 8% according to the question
Interest can be calculated by subtracting the principal from the amount paid back = 530-500 = 30$
We now plug these values into the equation
30 = (500 × 8× T)/100
100 × 30 = 4000T
T = 3000/4000
T = 0.75 (same as 0.75 × 12 months = 9 months)
Step-by-step explanation:
Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.
Type ni xi si
Graded 42 0.486 0.187
No-fines 42 0.359 0.158
To find - a. Formulate the above in terms of a hypothesis testing problem.
b. Give the test statistic and its reference distribution (under the null hypothesis).
c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.
Proof -
a.)
Hypothesis testing problem :
H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
b)
Test statistic :




⇒Z(cal) = 3.3687
Z(tab) = 1.96
As Z (cal) > Z(tab)
So, we reject H0 at 5% Level of significance
p-value = 0.99962
Hence
There is significant difference in mean conductivity at the two materials.
Answer:
a) false
b) true
c) false
d) false
Step-by-step explanation:
a) p-value is compared with test statistic to either accept or ereject the null hypothesis. There is no fixed p-value to reject the null hypothesis
b) p-value tells us the probabiltiy of finding null hypothesis to be true
c) There is no fixed p-value for nullyfying the the null hypothesis
d) There is no fixed p-value to reject the null hypothesis