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Ratling [72]
3 years ago
13

Brainliest to the right answer!

Mathematics
2 answers:
marshall27 [118]3 years ago
7 0

Answer: 5 1/4

Step-by-step explanation:

Sergeu [11.5K]3 years ago
5 0

Answer:

Step-by-step explanation:

Its D

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A shepherd is paid rs.2400 for gazing 18 cows for 5 weeks.How much will be paid for gazing 20 cows for 8 weeks?
lys-0071 [83]

Answer:

Rs. 4266.67(approx) will be paid for gazing 20 cows for 8 weeks.

Explanation:

A shepherd is paid rs. 2400 for gazing 18 cows for 5 weeks .

⇒ Number of money paid for gazing 1 cow for 1 week is, \frac{2400}{18\times 5} =\frac{400}{15}= \frac{80}{3}

then,

for gazing 20 cows for 8 weeks is=  Rs. \frac{80}{3}\times (20\times 8)

Therefore, the amount paid for gazing 20 cows for 8 weeks is, \frac{80 \times 160}{3} = \frac{12800}{3} = Rs. 4266.66667 or Rs. 4266.67(approx)

7 0
3 years ago
1. Flight 202's arrival time is normally distributed with a mean arrival time of 4:30
Bezzdna [24]

Answer:

Step-by-step explanation:

1) Let the random time variable, X = 45min; mean, ∪ = 30min; standard deviation, α = 15min

By comparing P(0 ≤ Z ≤ 30)

P(Z ≤ X - ∪/α) = P(Z ≤ 45 - 30/15) = P( Z ≤ 1)

Using Table

P(0 ≤ Z ≤ 1) = 0.3413

P(Z > 1) = (0.5 - 0.3413) = 0.1537

∴ P(Z > 45) = 0.1537

2)  By compering (0 ≤ Z ≤ 15) ( that is 4:15pm)

P(Z ≤ 15 - 30/15) = P(Z ≤ -1)

Using Table

P(-1 ≤ Z ≤ 0) = 0.3413

P(Z < 1) = (0.5 - 0.3413) = 0.1587

∴ P(Z < 15) = 0.1587

3) By comparing P(0 ≤ Z ≤ 60) (that is for 5:00pm)

P(Z ≤ 60 - 30/15) = P(Z ≤ 2)

Using Table

P(0 ≤ Z ≤ 1) = 0.4772

P(Z > 1) = (0.5 - 0.4772) = 0.0228

∴ P(Z > 60) = 0.0228

6 0
3 years ago
The base of a triangle is twice its hypotenuse, the sine of the angle opposite to its base is
svetoff [14.1K]
The answer   b= 2h
the angle  opposite to its base is a
so sina = b/2h= b/b=1
the answer is <span>b) 1</span>
6 0
3 years ago
Read 2 more answers
5379 divided by seven
Nookie1986 [14]
It is 768.4285714285714
8 0
3 years ago
Jadyn prepares a bag of candy before a road trip. The bag contains individual pieces of Skittles, M&amp;Ms, Life Savers and Sour
kipiarov [429]

Given condition is- The bag contains individual pieces of Skittles, M&Ms, Life Savers and Sour Patch Kids.

Let the total number of candies in the bag be = x

Then, Skittles are \frac{1x}{2}

Candies left = x-\frac{x}{2} = \frac{x}{2}

M&Ms are = \frac{1x}{8}

Now remaining candies are= \frac{3x}{8}

Now Life savers are = \frac{9x}{32}

Now adding all these 3 candies

\frac{x}{2}+ \frac{x}{8}+\frac{9x}{32}

=\frac{29x}{32}

Now the remaining candies sour patch will be =

x-\frac{29x}{32} =\frac{3x}{32}

Hence, sour patch kids are \frac{3}{32}

3 0
3 years ago
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