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Softa [21]
3 years ago
6

Solve for b −3b+2.5=4 Two step equations with fractions and decimals

Mathematics
2 answers:
jenyasd209 [6]3 years ago
4 0
-3b+2,5=4
-3b=4-2.5
-3b=1.5
b=1.5/(-3)
b= -0.5

DIA [1.3K]3 years ago
3 0
−3b+2.5=4

Subtract 2.5 from both sides.

−3b+2.5−2.5=4−2.5

−3b=1.5

Divide both sides by -3.

−3b/−3 = 1.5/−3

b=−0.5

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3 years ago
impeded by the current the outing club took 4 hours and 24 minutes to paddle 11 km up the exeter river to their campsite last we
lilavasa [31]

Answer: if there had been no current, the rate would have been 4 km/hr.

The rate of the current is 1.5km/hr

Step-by-step explanation:

Let x represent the speed of the boat in still water.

Let y represent the speed of the current.

Impeded by the current the outing club took 4 hours and 24 minutes(4 + 24/60 = 4.4 hours) to paddle 11 km up the exeter river to their campsite last weekend. This means that the total speed with which they travelled is (x - y) km/h

Distance = speed × time. Therefore

11 = 4.4(x - y)

Dividing through by 4.4, it becomes

2.5 = x - y - - - - - - - - - - -1

The next day the same current was with them and it took only 2 hours to make the return trip to campus. This means that the total speed with which they travelled is (x + y) km/h. The distance travelled is

11 = 2(x + y)

Dividing through by 2, it becomes

5.5 = x + y - - - - - - - - - - -2

Adding equation 1 to equation 2, it becomes

8 = 2x

x = 8/2 = 4

Substituting x = 4 into equation 1, it becomes

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3 0
3 years ago
In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In ad
zysi [14]

Answer:  a) 0.0792   b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)   (1)

Similarly ,

P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

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