Many inequalitys have -25 in their solution sets.
There are too many to name.
Answer:
There were 7 cats
Step-by-step explanation:
In this question, we are asked to calculate the number of individual cats and dogs present at an animal shelter on a certain Monday.
Since we do not know the exact numbers, let’s represent these animals by variables. Let the number of dogs be d and the number of cats be c .
Since there are a total of 9 animals, this means;
c + d = 9 •••••••••(I)
Let’s now get the second equation out of their finances;
Total amount spent on dogs will be 7 * d = 7d while total amount spent on cats will be 5.5 * c = 5.5c. If we add both, we will get the total amount spent which is $52.5
Mathematically therefore;
5.5c + 7d = 52.5 ••••••••(ii)
Now from 1, we can say that c = 9 - d
Let’s substitute this in ii
5.5(9-d) + 7d = 52.5
49.5 - 5.5d + 7d = 52.5
1.5d = 52.5 - 49.5
1.5d = 3
d = 3/1.5 = 2
Since d = 2, c = 9 - d = 9 - 2 = 7
Answer:
The critical value of <em>t</em> is 2.000.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean in case the population standard deviation is not known is:
The information provided is:
<em>n</em> = 57
Confidence level = 95%
As the sample size is large enough, i.e. <em>n</em> = 57 > 30 the sampling distribution of sample mean can be approximated by the normal distribution.
The distribution of sample statistic is normal and the sample size is quite large. This implies that it is appropriate to use a <em>t</em>-interval.
Compute the critical value of <em>t</em> as follows:
*Use a <em>t</em>-table for the critical value.
Thus, the critical value of <em>t</em> is 2.000.
Step 1: 10 grpups of 23 = 230
Step 2: 6 groups of 23 = 138
Step 3: 10 + 6 = 16