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finlep [7]
3 years ago
12

The ratio of rainy days to non rainy days was 2 to 3. How many of the 30 days in April were rainy?

Mathematics
2 answers:
9966 [12]3 years ago
4 0
(30/(2+3))2=12

So there were 12 rainy days in April.

check...

If there are 2:3, rainy to non-rainy there are 5 equal parts of magnitude:  30/5=6

So each part is equal to 6.  So 6*2=12 rainy days and 6*3=18 non-rainy days.
viktelen [127]3 years ago
3 0
20 days were rainy.

2:3=x:30
2/3=x/30
x=20
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slamgirl [31]

Here,

2x^2 + 7x + 1 = 0

given,

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From formula,

x = (-b+-√b^2-4ac) / 2a.

We get,

x = -7/4 ± √(41)/4 Ans.

4 0
3 years ago
Find the value of x, y and z. Write reasons
WARRIOR [948]

Answer:

z= 115 (vertically opposite angles)

x= 65 (straight angle) 180-115=65

y=65 (vertically opposite angle)

Step-by-step explanation:

7 0
3 years ago
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The Scott family is trying to save as much money as possible. One way to cut back on the money they spend is by finding deals wh
avanturin [10]

Answer:

Unit price of strawberries at Grocery Mart is $ 1.495 or 150 pennies

Unit price of strawberries at Baldwin Hills Market is $ 1.33 or 133 pennies.

Step-by-step explanation:

1 dollar = 100 pennies

Given:

Cost of 2 pounds of strawberries at Grocery Mart = $ 2.99

Cost of 3 pounds of strawberries at Baldwin Hills Market = $3.99

∵ Cost of 2 pounds of strawberries at Grocery Mart = $ 2.99

∴ Cost of 1 pound of strawberries at Grocery Mart = \frac{2.99}{2}=\$1.495=1.495\times 100\approx 150\textrm{ pennies}

∵ Cost of 3 pounds of strawberries at Baldwin Hills Market = $3.99

Cost of 1 pound of strawberries at Baldwin Hills Market = \frac{3.99}{3}=\$1.33= 1.33\times 100 = 133\textrm{ pennies}

Therefore, the unit price of strawberries at each grocery store is the cost of 1 pound of strawberries. So, unit rate at Grocery Mart is 150 pennies and at Baldwin Hills Market is 133 pennies.

7 0
3 years ago
4(x+3)-4=8(1/2x+1) i need to find the solution
oee [108]
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5 0
4 years ago
Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose P(V) = 0.17
ArbitrLikvidat [17]

Answer: 0.82

Step-by-step explanation:

The probability of the computer not containing neither a virus nor a worm is expressed as P(V^{C}∩W^{C}) , where P(V^{C}) is the probability that the event V doesn't happen and P(W^{C}) is the probability that the event W doesn't happen.

P(V^{C})= 1-P(V) = 1-0.17 = 0.83

P(W^{C})=1-P(W) = 1-0.05 = 0.95

Since V^{C} and W^{C} aren't mutually exclusive events, then:

P(V^{C}∪W^{C}) = P(V^{C}) + P(W^{C}) - P(V^{C}∩W^{C})

Isolating the probability that interests us:

P(V^{C}∩W^{C})= P(V^{C}) + P(W^{C}) - P(V^{C}∪W^{C})

Where P(V^{C}∪W^{C}) = 1 - 0.04 = 0.96

Finally:

P(V^{C}∩W^{C}) = 0.83 + 0.95 - 0.96 = 0.82

5 0
3 years ago
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