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frez [133]
2 years ago
12

A recipe for cookies calls for $5 of a cup of brown sugar and makes 6 cookies. If you adjust the recipe to make 8 cookies, how m

uch brown sugar will you need?​
Mathematics
1 answer:
Klio2033 [76]2 years ago
5 0
Ur not that slow just try
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Please help me I need someone to help me with this ​
Digiron [165]

Answer:

-2

Step-by-step explanation:

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6 0
2 years ago
Read 2 more answers
14/15 divided by 7/5
Charra [1.4K]

9514 1404 393

Answer:

  2/3

Step-by-step explanation:

There are a couple of different ways that division of fractions can be done.

1) "invert and multiply"

  \dfrac{14}{15}\div\dfrac{7}{5}=\dfrac{14}{15}\times\dfrac{5}{7}\\\\=\dfrac{14}{7}\times\dfrac{5}{15}=\boxed{\dfrac{2}{3}}

__

2) use the numerators when the denominators are the same

  \dfrac{14}{15}\div\dfrac{7}{5}=\dfrac{14}{15}\div\dfrac{21}{15}\\\\=\dfrac{14}{21}=\boxed{\dfrac{2}{3}}

6 0
3 years ago
The perimeter of the big wheel of a bicycle is 9 cm longer than the perimeter of the front wheel. On the road of 120 cm, the big
beks73 [17]

Answer:

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Question

Bookmark

A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60cm, calculate the speed per hour with which the boy is cycling.

Easy

Solution

verified

Verified by Toppr

Radius of the wheel =r=

2

60

=30 cm

Circumference of the wheel =2πr=2×

7

22

×30=

7

1320

cm

Distance covered in 1 revolution =

7

1320

cm

Distance covered in 140 revolutions =

7

1320

×140=26400 cm=264 m=

1000

264

km

Distance covered in one minute = Distance covered in 140 revolutions =

1000

264

km

Distance covered in 1 hour =

1000

264

×60=15.84 km

Hence, the speed with which the boy is cycling is 15.84km/hr.

3 0
2 years ago
Solving Systems of Equations by Graphing - Scavenger Hunt
Ostrovityanka [42]
Where’s the question
3 0
3 years ago
A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy
aleksandrvk [35]

Answer:

(a) A y P(A) = 0.4 (b) \bar{B} y P(\bar{B})=0.5 (c) \bar{A}∪\bar{B} y P(\bar{A}∪\bar{B}) = 0.9 (d) \bar{A}∩\bar{B} y P(\bar{A}∩\bar{B})=0.2

Step-by-step explanation:

A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So

P(A\cap \bar{B}) = 0.3, P(B\cap \bar{A}) = 0.4 and P(A\cap B) = 0.1

(a) P(A) = P(A\cap (B\cup\bar{B})) = P(A\cap B) + P(A\cap \bar{B}) = 0.1 + 0.3 = 0.4

(b)  P(B) = P(B\cap (A\cup\bar{A})) = P(B\cap A) + P(B\cap \bar{A}) = 0.1 + 0.4 = 0.5

P( \bar{B}) = 1-P(B) = 1-0.5 = 0.5

(c) The customer does not buy from outlet 1 is the complement of A, i.e.,  \bar{A}, and the customer does not buy from outlet 2 is the complement of B, i.e.,  \bar{B}, so, the customer does not buy from outlet 1 or does not buy from outlet 2 is  \bar{A}∪ \bar{B} and P(\bar{A}∪ \bar{B}) = P((A\cap B)^{c}) by De Morgan's laws

P((A\cap B)^{c})  = 1-P(A∩B)=1-0.1=0.9

(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to \bar{A}∩\bar{B} and P( \bar{A}∩\bar{B}) = P((AUB)^{c}) by De Morgan's laws, and

P((AUB)^{c}) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2

8 0
2 years ago
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