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slavikrds [6]
3 years ago
7

The histogram shows the ages of the volunteers who cleaned up a beach.

Mathematics
2 answers:
statuscvo [17]3 years ago
6 0
<h2>Answer:</h2>

The percent of the volunteers who were at least 15 years old but no more than 24 years old are:

                            30%

<h2>Step-by-step explanation:</h2>

We will make a table based on the histogram:

    Age-group            Number of students

       10-14                               3

       15-19                                2

       20-24                              7

       25-29                              8

       30-34                              5

       35-39                              5

Hence, the number of volunteers who are atleast 15 years no more than 24 years old are those who come in the age-group 15-19 and 20-24.

Hence, total number of such students= 2+7=9

and the total number of students= 3+2+7+8+5+5= 30

Hence, the percent of volunteers who lie between 15-24 years of age are:

=\dfrac{9}{30}\times 100\\\\=30\%

                    Hence, the answer is: 30%

Greeley [361]3 years ago
3 0
The answer is 29.03%. because there are 2 people who are 15-19 and there are 7 people between 20-24. 7+2=9. and there are a total of 31 people. so 9 out 31 as a percent is 29.03%. hope this helps. please give me a brainliest
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Find the 8th term of the geometric sequence 5, -15, 45
FrozenT [24]

Answer:

a₈ = - 10935

Step-by-step explanation:

the nth term of a geometric sequence is

a_{n} = a₁ (r)^{n-1}

where a₁ is the first term and r the common ratio

here a₁ = 5 and r = \frac{a_{2} }{a_{1} } = \frac{-15}{5} = - 3 , then

a₈ = 5 × (-3)^{7} = 5 × - 2187 = - 10935

6 0
2 years ago
3.3^(2x+1)-103^x+1=0 need value of x
photoshop1234 [79]

Answer:

The value of x is approximately -1.531.

Step-by-step explanation:

Let 3.3^{2\cdot x + 1}-103^{x+1} = 0, we proceed to solve this expression by algebraic means:

1) 3.3^{2\cdot x + 1}-103^{x+1} = 0  Given

2) 3.3^{2\cdot x}\cdot 3.3 -103^{x}\cdot 103 = 0 a^{b}\cdot a^{c} = a^{b+c}

3) (3.3^{x})^{2}\cdot 3.3 -\left[\left( \sqrt{103} \right)^{2}\right]^{x}\cdot 103 = 0 (a^{b})^{c} = a^{b\cdot c}

4) (3.3^{x})^{2}\cdot 3.3 - \left[\left(\sqrt{103}\right)^{x}\right]^{2}\cdot 103 = 0 (a^{b})^{c} = a^{b\cdot c}/Commutative property

5) \left[\left(\frac{3.3}{\sqrt{103}}\right)^{x}\right] ^{2}-\frac{103}{3.3} = 0 Existence of multiplicative inverse/Definition of division/Modulative property/a^{b}\cdot a^{c} = a^{b+c}

6) \left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\frac{103}{3.3} Existence of additive inverse/Modulative property/(a^{b})^{c} = a^{b\cdot c}

7) \log \left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\log \frac{103}{3.3} Definition of logarithm.

8) 2\cdot x\cdot \log \left(\frac{3.3}{\sqrt{103}} \right)= \log \frac{103}{3.3}     \log_{b} a^{c} = c\cdot \log_{b} a

9) 2\cdot x \cdot [\log 3.3-\log \sqrt{103}] = \log 103 - \log 3.3      \log_{b} \frac{a}{d}

10) x\cdot (2\cdot \log 3.3-\log 103) = \log 103 - \log 3.3     \log_{b} a^{c} = c\cdot \log_{b} a/Associative property

11) x = \frac{\log 103-\log 3.3}{2\cdot \log 3.3-\log 103}   Existence of multiplicative inverse/Definition of division/Modulative property

12) x \approx -1.531  Result

The value of x is approximately -1.531.

4 0
3 years ago
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Veronika [31]
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8 0
4 years ago
Have NO EXPLANATION Than Don't ANSWER
kykrilka [37]

Answer:

D

Step-by-step explanation:

Using the Distributive properties.

that one is the simplified one and the regular problem.

5 0
3 years ago
Please help!<br><br> -1 1/3 ÷ 2 2/5=
muminat
The answer to this equation is -1/15
7 0
2 years ago
Read 2 more answers
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