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igor_vitrenko [27]
3 years ago
13

in a recent year, the weather was partly cloudy 2/5 of the days. Assuming there are 365 days ina year, how many days were partly

cloudy?
Mathematics
2 answers:
andrew-mc [135]3 years ago
8 0

Answer:

146 days.

Step-by-step explanation:

We have been given that in a recent year, the weather was partly cloudy 2/5 of the days. We are asked to find the number of days were party cloudy in a year.

We know that one year equals 365 days. To find the number of party cloudy days in a year, we need to find 2/5 of 365.

\text{Number of party cloudy days in a year}=\frac{2}{5}\times 365

\text{Number of party cloudy days in a year}=2\times 73

\text{Number of party cloudy days in a year}=146

Therefore, 146 days in a year were partly cloudy.

Karolina [17]3 years ago
3 0


Here's two ways

1. (365*2)/5 =

730/5 =

146 days

2. 365/5=

73*2

146 days

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The given compound inequality is

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First we solve the first inequality,

2(x+3)>6
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Now we solve the second inequality

2x+3\leq -7
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So we have

x > 0 \ or  \ x\leq  -5

So the required solution is

( - \infty,-5] \ or \ (0, \infty)

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3 years ago
What is 8.25 rounded off to the nearest whole number
Ugo [173]

Answer:

8

Step-by-step explanation:.25 is lower than .5

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Because gambling is a big​ business, calculating the odds of a gambler winning or losing in every game is crucial to the financi
IRISSAK [1]

Answer:

a) 0.1165

b) 0.0983

c) 0.000455

d) 0.787

e) 0.767

Step-by-step explanation:

5 ​bars, 4 ​lemons, 3 ​cherries, and a bell

Total = 5+4+3+1 = 13

The probability of getting a bar on a slot, P(Ba) = 5/13 = 0.385

A lemon, P(L) = 4/13 = 0.308

A cherry, P(C) = 3/13 = 0.231

A bell, P(Be) = 1/13 = 0.0769

a) Probability of getting 3 lemons = (4/13) × (4/13) × (4/13) = 256/2197 = 0.1165

b) Probability of getting no fruit symbol

On each slot, there are 4+3 = 7 fruit symbols.

Probability of getting a fruit symbol On a slot = 7/13

Probability of not getting a fruit symbol = 1 - (7/13) = 6/13 = 0.462

Probability of not getting a fruit symbol On the three slots = 0.462 × 0.462 × 0.462 = 0.0983

c) Probability of getting 3 bells, the jackpot = (1/13) × (1/13) × (1/13) = 1/2197 = 0.000455

d) Probability of not getting a bell on the 3 slots

Probability of not getting a bell on one slot = 1 - (1/13) = 12/13 = 0.923

Probability of not getting a bell on the 3 slots = (12/13) × (12/13) × (12/13) = 1728/2197 = 0.787

e) Probability of at least one bar is a sum of probabilities

Note that Probability of getting a bar = 5/13 and probability of not getting a bar = 8/13

1) Probability of getting 1 bar and other stuff on the 2 other slots (this can happen in 3 different orders) = 3 × (5/13)×(8/13)×(8/13) = 960/2197 = 0.437

2) Probability of getting 2 bars and other stuff on the remaining slot (this can also occur in 3 different orders) = 3 × (5/13)×(5/13)×(8/13) = 600/2197 = 0.273

3) Probability of getting 3 bars on the slots machine = (5/13) × (5/13) × (5/13) = 125/2197 = 0.0569

Probability of at least one bar = 0.437 + 0.273 + 0.0569 = 0.7669 = 0.767

5 0
3 years ago
Sherry was in charge of distributing 25 food items that were donated to the local food pantry. on moday she distributed 8 items
creativ13 [48]

Answer:

Step-by-step explanation:

First you add up all the numbers - 8+7+5=20

Then you know 5 is 1/5 of 25

You know 5+5+5+5=20

ANSWER - 4/5

3 0
2 years ago
Sarah is a computer engineer and manager and works for a software company. She receives a
daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = a

Given that:

a_4=129\\a_{10}=207

Formula for n^{th} term of an Arithmetic Progression is given as:

a_n=a+(n-1)d

Where d will represent the number of projects increased every year.

and n is the year number.

a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)

Subtracting (2) from (1):

78 = 6d\\\Rightarrow d =13

By equation (1):

129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90

<em>Number of projects in the first year = 90</em>

<em></em>

<em>b) </em>

Number of projects in the twelfth year =

a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233

Each project pays $500

Earnings in the twelfth year = 233 \times 500 = $116500

Sum of an AP is given as:

S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 \times 1938 = $969000

8 0
2 years ago
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