This is a system of equations problem, so we need two equations to represent each case:
5s+5L=155
10s+12L=346
Now we isolate one of the variables.
5s = 155 - 5L
s = 31 - L
Now we substitute s in the other equation with this equation.
10(31-L) + 12L = 346
310-10L + 12L = 346
2L = 36
L = $18
Now to find s we need to plug L into one of the equations.
10s + 12(18) = 346
10s + 216 = 346
10s = 130
s = $13
A large box costs$18 and a small box costs $13
Hope this helps.
The largest possible last digit in the string of 2002 digits and number divisible by 19 or 31 is 9.
Given the first digit of a string of 2002 digits is 1 and the two digit number formed by consecutive digits within the string is divisible by 19 or 31.
We have to tell the last largest digit of such number.
Two digit numbers divisible by 19=19,38,57,76,95.
Two digit numbers divisible by 31=31,62,93,124
Number started with 1 =19
Last digit is 9
We have said that the number should be divisible by 19 or 31 not from both and started with 1.
Hence the largest possible last digit and number divisible by 19 or 31 in this string is 9.
Learn more about digits at brainly.com/question/26856218
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Answer:
The test statistic for this test is 3.82.
Step-by-step explanation:
The null hypothesis is:
The alternate hypotesis is:
Our test statistic is:
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation of the population and n is the size of the sample.
In this question:

So
The test statistic for this test is 3.82.
The answer is 70 the third option.