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slamgirl [31]
3 years ago
5

C. what percent of the days were sunny or rainy?

Mathematics
2 answers:
Oxana [17]3 years ago
7 0
Luckily, all of the denominators are factors of 100.
rainy: 12/100
sunny: 44/100
partly cloudy: 30/100
cloudy: 10/100
happy learning!
-mathgurl
tresset_1 [31]3 years ago
4 0

Answer:

(c) 60% of the days were sunny or rainy.

(d) 40% of the days were cloudy or partly cloudy.

Step-by-step explanation:

We have been given a pie chart, which represent weather during September.

(c). To find the percent of the days that are sunny or rainy, we will add sunny days to rainy days and then convert them into percent as:

\text{Days that are sunny or rainy}=\frac{4}{25}+\frac{11}{25}

\text{Days that are sunny or rainy}=\frac{4+11}{25}

\text{Days that are sunny or rainy}=\frac{15}{25}

\text{Percent of the days that are sunny or rainy}=\frac{15}{25}\times 100

\text{Percent of the days that are sunny or rainy}=15\times 4

\text{Percent of the days that are sunny or rainy}=60

Therefore, 60% of the days were sunny or rainy.

(d). To find the percent of the days that are cloudy or partly cloudy, we will add cloudy days to partly cloudy days and then convert them into percent as:

\text{Days that are cloudy or partly cloudy}=\frac{1}{10}+\frac{3}{10}

\text{Days that are cloudy or partly cloudy}=\frac{1+3}{10}

\text{Days that are cloudy or partly cloudy}=\frac{4}{10}

\text{Percent of the days that are cloudy or partly cloudy}=\frac{4}{10}\times 100

\text{Percent of the days that are cloudy or partly cloudy}=4\times 10

\text{Percent of the days that are cloudy or partly cloudy}=40

Therefore, 40% of the days were cloudy or partly cloudy.

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Use the graph of f(x) = x2<br> to write an equation for the function represented by each graph.
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4 0
3 years ago
QUESTION 11.1
lawyer [7]

The group paid $ 5250 at first city and $ 6250 at second city

<u>Solution:</u>

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the  second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the  tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

<em><u>Therefore, a equation is framed as:</u></em>

8.5 % of x + 5.5 % of y = 790

\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790

0.085x + 0.055y = 790 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

<h3>x = 5250</h3>

<em><u>Substitute x = 5250 in eqn 1</u></em>

y = 5250 + 1000

<h3>y = 6250</h3>

Thus the group paid $ 5250 at first city and $ 6250 at second city

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