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Inga [223]
3 years ago
12

Can somebody help me please ​

Mathematics
1 answer:
julsineya [31]3 years ago
4 0

Answer:

3x+2y=18.65

Step-by-step explanation:

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What in 40 023 032 in word form??
krok68 [10]
Forty million, twenty-three thousand and thirty-two
5 0
3 years ago
Determine the measure of ∠Y.
ira [324]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
A researcher wishes to determine whether there is a difference in the average age of middle school, high school, and college tea
MakcuM [25]

The correct statement that can be made about the means is a. There is not enough evidence to suggest that the means are different.

<h3>What is true about the means?</h3><h3 />

Given that α = 0.01, we can use an ANOVA analysis to determine the p-value of the means.

When we run the means through an ANOVA software, the p-value can be found to be 0.1142.

This figure is greater than α = 0.01.

This means that we do not have the evidence to reject the null hypothesis that the means are different.

Options for this question:

  • a. There is not enough evidence to suggest that the means are different.
  • b. The mean age of middle school teachers is different from the mean age of high school teachers.
  • c. The mean age of middle school teachers is different from the mean age of college teachers.
  • d. The mean age of high school teachers is different from the mean age of college teachers.

Find out more on the p-value at brainly.com/question/4621112

#SPJ1

7 0
2 years ago
There are 4512 marbles which will be divided equally among 48 students how many marbles will each student get
Grace [21]

Answer:

94 marbles

Step-by-step explanation:

4512/48 = 94

3 0
3 years ago
Help pls I'll give 25 points
RideAnS [48]

Answer:

\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are 3!=6 ways to rearrange the remaining 3 digits for a total of 2\cdot 6=\boxed{12} possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still 3!=6 ways to rearrange the remaining three numbers. Therefore, there are 1\cdot 6=\boxed{6} possible numbers when 1 is the fourth digit of the number.

Thus, there are 24+18+12+6=\boxed{60} numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

5 0
3 years ago
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