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Serggg [28]
3 years ago
6

During February, Kevin will water his ivy every third day, and water his cactus every fifth day. On which date will Kevin first

water both plants together? Will Kevin water both plants together again in February? Explain.
Mathematics
1 answer:
frutty [35]3 years ago
4 0
Kevin will water his plants on the 15th day. He wont water his plants again in February because there are only 28 days in that month, and even on a leap year, there would still be only 29 days
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How to convert 9.4 into a fraction
True [87]
9.4 =  \frac{9.4}{1} = \frac{9.4\times 10}{1 \times 10} = \frac{94}{10} = \frac{47 \times 2}{5 \times 2} =\frac{47}{5} = \frac{45+2}{5}= \frac{45}{5}   + \frac{2}{5} =9 \frac{2}{5}
5 0
3 years ago
Read 2 more answers
A test taker gets 70 on 1st exam, 80 on 2nd exam, 2/3 of 4/5 of his 2nd exam on his 3rd test. If the professor gives 5 points ex
Serga [27]
<h3>Answer:  127</h3>

=================================================

Explanation:

The phrasing "2/3 of 4/5 of his 2nd exam on his 3rd test" is a bit clunky in my opinion. It seems more complicated than it has to be.

The student got 80 on the second exam. 4/5 of this is (4/5)*80 = 0.8*80 = 64. Then we take 2/3 of this to get (2/3)*64 = 42.667 approximately. If we assume only whole number scores are given, then this would round to 43.

Let x be the score on the fourth exam. Since 5 points of extra credit are given, the student actually got x+5 points on this exam.

So we have these scores

  • first exam = 70
  • second exam = 80
  • third exam = 43
  • fourth exam = x+5

Adding up these scores and dividing by 4 will get us the average

(sum of scores)/(number of scores) = average

(70+80+43+x+5)/4 = 80

(x+198)/4 = 80

x+198 = 4*80

x+198 = 320

x = 320 - 198

x = 122

So the student got a score of x+5 = 122+5 = 127 on the fourth exam.

7 0
3 years ago
50 POINTS !!<br><br><br> PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.
Firdavs [7]

Answer:

According to hypotenuse

39²+b²=65²

b²=65²-39²

b²=4225-1521

b=√2,704

b=52 km

3 0
3 years ago
Read 2 more answers
Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial t
jekas [21]

Answer:

(12.1409, 14.0591

Step-by-step explanation:

Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.

Group   Group One     Group Two  

Mean 87.600 74.500

SD 1.000 1.500

SEM 0.316 0.433

N 10      12    

The mean of Group One minus Group Two equals 13.100

standard error of difference = 0.556

 90% confidence interval of this difference:  

(13.1-1.725*0.556,13.1+1.725*0.556)\\=(12.1409, 14.0591)

  t = 23.5520

 df = 20

4 0
3 years ago
9n - 2 - 6n - 3<br> I need help with question.
andrew11 [14]
9n - 2 - 6n - 3
(9n + -6n) + (-2+-3)
=3n -5
8 0
4 years ago
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