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Damm [24]
2 years ago
6

A carpenter bought a piece of wood that was 6.85 meters long. Then she sawed 0.57 meters off the end. How long is the piece of w

ood now? meters. help plz​
Mathematics
1 answer:
andrey2020 [161]2 years ago
8 0

Answer:

6.28meters

it is the aswer if wrong please tell me

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The Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
kirza4 [7]

Answer:

Step-by-step explanation:

given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging

Clerk          I           II           III    total

   

Pack      0.33          0.23    0.44 1

   

Defective 0.02 0.025    0.015  

   

Pack&def 0.0066 0.00575 0.0066 0.01895

a)  probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate

= P(II clerk) -P(II clerk and defective) = 0.23-0.00575=0.22425

b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)

=0.01895

c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3

=P(clerk 3 and def)/P(defective)

=\frac{0.0066}{0.01895} \\=0.348285

8 0
3 years ago
The point A(-6, 8) has been transformed using the composition r(90,O) counterclockwise ∘Rx−axis . Where is A'?
Vikki [24]
A would now be a positive 6
7 0
2 years ago
Leah has $4.35 in her pocket in nickels and dimes. The number of dimes is 2 less than 3 times the number of nickels. How many of
Maksim231197 [3]
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6 0
2 years ago
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An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then
Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

\angle Q=110^\circ

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)

(b)Flight Direction of Y

Using Law of Sines

\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)

The direction of flight Y to the nearest degree is 39 degrees.

7 0
3 years ago
What is 44/154 simplified to
svlad2 [7]

Answer: 2/7

Both 44 and 154 have the greatest common factor of 22.

44/22 = 2

154/22 = 7

5 0
3 years ago
Read 2 more answers
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