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Naya [18.7K]
3 years ago
5

When choosing a new class color, 90 students voted for crimson and 60 voted for royal blue. What percent of these students voted

for the color crimson?
Mathematics
1 answer:
neonofarm [45]3 years ago
6 0
The percentage of students voted for colour crimson
=
\frac{90}{60 + 90} \times 100\%  \\  =  \frac{90}{150}  \times 100\%\\  =  \frac{9}{15} \times 100\% \\  = 60\%
hope it helps!
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Please help !<br> I need the answers for #43 and #44
Vlad [161]

43. 9 + 10 + 13 + 13 = $45

45 - 1 = 44

44 x 4 = 176

176 + 8 = 184

184/4 = 46

46 tokens per person

44. a. 22 x 5 = 110

943/110 = 8.7

you will need 9 bookcases

b. 110 x 9 = 990

990 - 943 = 47

47 - 44 = 3

There will be 3 books on the third shelf.

4 0
3 years ago
A store is having a sale on hats . The sale is buy one hat , get one 50% off . The original price of one is 24.98 . Alex bought
Alex73 [517]

50% of 24.98 = $12.49

24.98 + 12.49 = 37.47

187.34 / 37.47 = 5

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3 0
3 years ago
18) The degree of P(x) = (x-1)(x+2)(x+3)(2x-5) is<br><br><br>4<br>3<br>2<br>1​
Fantom [35]

Answer:

degree 4

Step-by-step explanation:

The degree of a polynomial is determined by the term with the largest exponent for the polynomial in standard form

Given

P(x) = (x - 1)(x + 2)(x + 3)(2x - 5)

We need only consider the product of the leading terms in each factor, that is

x(x)(x)(2x) = 2x^{4} ← is the leading term in the expansion of the factors

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3 0
2 years ago
Anybody help me please!!!!!!!!
SCORPION-xisa [38]

Answer:

A reflection over the x-axis

Step-by-step explanation:

7 0
3 years ago
You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%
bezimeni [28]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}

\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}

6 0
3 years ago
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