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Crank
2 years ago
7

A homeowner puts a passcode-enabled lock on her front door. To choose a passcode, she must choose a number, a letter from a list

of 5 letters, and then another number.
Number Letter Number
0 A 0
1 E 1
2 I 2
3 O 3
4 U 4
5 5
6 6
7 7
8 8
9 9
How many possible passcodes can she make?
Mathematics
1 answer:
Daniel [21]2 years ago
4 0

Answer:

500

Step-by-step explanation:

I assume each number is from 0 to 9. Also there are 5 letters.

10 * 5 * 10 = 500

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If the average (arithmetic mean) of 2,7, and x is 12, what is the value of x
guajiro [1.7K]

x is equal to 27.

In order to find the value of x, you have to use the average formula and solve for the missing variable. The mean formula is all of the numbers added together divided by the number of inputs.

(inputs added)/(number of inputs) = Average

(2 + 7 + x)/3 = 12 ----> Multiply by 3

9 + x = 36 ----> Subtract 9

x = 27

3 0
2 years ago
During a test period, an experimental group of 10 vehicles using an 85 percent ethanol-gasoline mixture showed mean CO2 emission
PIT_PIT [208]

Answer:

A quick comparison of the sample variances suggests that the population variances <u><em>are same or almost equal.</em></u>

Step-by-step explanation:

The populations from which two samples are drawn are normally distributed.

The  variable used to predict another variable is called an independent variable.

F- distribution is used for the test of  two sample hypotheses of variances.

If two populations have equal variances then the test Statistic is nearly equal to 1.

<em>t= s²₁/s²₂i</em>

<em />

As the variances are almost or nearly equal the test statistic would be almost equal to 1.

7 0
2 years ago
Ariel incorrectly says that 2 5/8 is the same as 2.58
Nata [24]

Answer:

ariel is incorrect because 2 5/8 is equal to 2.63

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Jake's morning run was a distance of 9.3 miles. He ran back and forth on one long straight street. He started at his house and r
Sergeeva-Olga [200]

Answer: the distances can be 0.65 miles (in the opposite direction to the school) or 4.65 miles (in the same direction as the school)

Step-by-step explanation:

Ok, the data that we have is:

Total distance = 9.3 mi.

The travel is:

House to school = 4 mi.

school to library = A

library to house = B

Now, we have that:

4mi + A + B = 9.3mi.

We have three possibilities:

1) The order of locations is: house, library, school

The travel from: school to library + library to house is equivalent to a travel between the school to the house = 4mi.

Then we have A + B = 4mi

4mi + A + B = 8mi ≠ 9.3mi

Then the library can not be between the house and the school.

2) The order of locations is: house, school, library.

In this case we have that the distance between the library and the house is equal to the distance between the house and the school plus the distance between the school and the library, then:

4mi + A = B.

We can replace this in our original equation:

4mi + A + B = 9.3mi

4mi  + A + (4mi + A) = 9.3mi

8mi + 2*A = 9.3mi

2*A = 9.3mi - 8mi = 1.3mi

A = 1.3mi/2 = 0.65mi

Then the distance between the house and the library is:

The 4 miles between the house and the school, plus the 0.65 miles between the school and the library:

Distance = 4mi + 0.65mi. = 4,65mi

3) The third case is when the order of the locations is:

Library, house, school.

Then the distance between the house and the library is equal to the distance between the school and the library minus the distance between the house and the school, this is:

A - 4mi = B

Now we can replace this in our distance equation:

4mi + A + B = 9.3mi

4mi + A + (A - 4mi) = 9.3 mi

2A = 9.3mi

A = 9.3mi/2 = 4.65mi

Then the distance between the house and the library is:

B = A - 4mi = 4.65mi - A = 0.65mi

Then the distance between the house and the library is 0.65 miles in this case.

7 0
2 years ago
Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw
PolarNik [594]

Answer:

Using <u>backward reasoning</u> we want to show that <em>"We can never get nine 0's"</em>.

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

<u>There were 9 </u><u>0's</u><u>:</u>

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

<u>There were 9 </u><u>1's</u><u>:</u>

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

<u />

5 0
2 years ago
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