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4 years ago

A tile installer has selected four different size square tiles to cover a floor. The areas of the tiles are:

Mathematics

1 answer:

snow_lady [41]4 years ago

3 0

Answer:

C and D

Step-by-step explanation:

The length of the sides will be irrational if it is not a whole number. The tile installer has a square shape, so the formula for area is

area= $side^{2}$

Since we need to find if the side has the whole number, we can make the function to solve for side

side= $\sqrt{area} }$

Then the side for each tile installer will be:

A= $\sqrt{25 in^{2} }$ = 5 inch

B= $\sqrt{36 in^{2} }$ = 6 inch

C= $\sqrt{46 in^{2} }$ = 6.78 inch

D= $\sqrt{60 in^{2} }$ = 7.74 inch

Both C and D have irrational sides

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Test scores: Scores on a statistics exam had a mean of 75 with a standard deviation of 5. Scores on a calculus exam had a mean o

beks73 [17]

Answer:

CV for statistics exam = 15%

CV for calculus exam = 19%

Since the CV for calculus exam is higher, it has a greater spread relative to the mean than the statistics exam.

Step-by-step explanation:

To find coefficient variation we use the formula:

CV = (SD/mean) * 100

CV for the statistics exam:

where; SD= 5

mean= 75

CV = ( 5/75) *100

= 0.15 or 15%

CV for calculus exam

SD = 11

Mean= 58

CV= (11 /58) * 100

= 0.19 or 19%

8 0

3 years ago

In a recent survey in a Statistics class, it was determined that only 70% of the students attend class on Fridays. From past dat

pishuonlain [190]

Answer:

a) 67.6% of students is expected to pass the course

b) 0.9112 = 91.12% probability that he/she attended classes on Fridays

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

a. What percentage of students is expected to pass the course?

88% of 70%(attended class)

20% of 100 - 70 = 30%(did not attend class). So

$p = 0.88*0.7 + 0.2*0.3 = 0.676$

0.676*100% = 67.6%

67.6% of students is expected to pass the course.

b. Given that a person passes the course, what is the probability that he/she attended classes on Fridays?

Here, we use conditional probability:

Event A: Passed the course

Event B: Attended classes on Fridays.

67.6% of students is expected to pass the course.

This means that $P(A) = 0.676$

Probability that passed and attended classes on Friday.

88% of 70%

This means that:

$P(A \cap B) = 0.88*0.7 = 0.616$

Then

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.616}{0.676} = 0.9112$

0.9112 = 91.12% probability that he/she attended classes on Fridays

5 0

3 years ago

Will the product of - 3x ^ 3 + 2x - 4 and x ^ 3 + x - 2 be the same as the product of x ^ 3 + x - 2 2 and - 3x ^ 3 + 2x - 4 Expl

Advocard [28]

Answer:s

Step-by-step explanation:

4 0

3 years ago

Find the difference.

Vsevolod [243]

$5.\\12\dfrac{3}{10}-7\dfrac{7}{10}=11\dfrac{10+3}{10}-7\dfrac{7}{10}=11\dfrac{13}{10}-7\dfrac{7}{10}=4\dfrac{6}{10}=4\dfrac{6:2}{10:2}=\boxed{5\dfrac{3}{5}}\\\\11-7=4\\\\\dfrac{13}{10}-\dfrac{7}{10}=\dfrac{13-7}{10}=\dfrac{6}{10}\\\\6.\\8\dfrac{1}{6}-3\dfrac{5}{6}=7\dfrac{6+1}{6}-3\dfrac{5}{6}=7\dfrac{7}{6}-3\dfrac{5}{6}=(7-3)+\dfrac{7-5}{6}=4\dfrac{2}{6}=4\dfrac{2:2}{6:2}=\boxed{4\dfrac{1}{3}}$

$9.\\7\dfrac{1}{6}-2\dfrac{5}{6}=6\dfrac{6+1}{6}-2\dfrac{5}{6}=6\dfrac{7}{6}-2\dfrac{5}{6}=(6-2)+\dfrac{7-5}{6}=4\dfrac{2}{6}=\boxed{4\dfrac{1}{3}}\\\\10.\\9\dfrac{3}{12}-4\dfrac{7}{12}=8\dfrac{12+3}{12}-4\dfrac{7}{12}=8\dfrac{15}{12}-4\dfrac{7}{12}=(8-4)+\dfrac{15-7}{12}=4\dfrac{8}{12}\\\\=4\dfrac{8:4}{12:4}=\boxed{4\dfrac{2}{3}}$

5 0

3 years ago

Please help me !!!!!! (wrong answers get reported)

Sholpan [36]

its D, I know it is. please don't report me

5 0

3 years ago