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3 years ago

0.6(4-2x)=20.5-(3x+10)

Mathematics

1 answer:

solong [7]3 years ago

5 0

The solution of 0.6(4-2x)=20.5-(3x+10) is x = 4.5

Solution:

Given, expression is 0.6 ( 4 – 2x ) = 20.5 – ( 3x + 10 )

We have to solve the above given equation for x.

Now, take the equation

$\rightarrow 0.6(4-2 x)=20.5-(3 x+10)$

We have to use BODMAS rule. BODMAS means sequence of operations which is brackets, order, division, multiplication, addition and then subtraction

On solving the brackets, we get

$\rightarrow 0.6 \times 4-0.6 \times 2 x=20.5-3 x-10$

Now multiplication is given preference

$\rightarrow 2.4-1.2 x=10.5-3 x$

$\rightarrow 3 x-1.2 x=10.5-2.4$

Now subtraction is done

$\rightarrow 1.8 \mathrm{x}=8.1$

Now division is performed

$\rightarrow x=\frac{8.1}{1.8}=4.5$

Hence, the value of x is 4.5

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3 years ago

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(b) The probability that at most three forms are required is 0.40.

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Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

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Use this law to compute the value of k.

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Thus, the value of k is $\frac{1}{15}$ .

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Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

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Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

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Check condition 1:

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Condition 1 is fulfilled.

Check condition 2:

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Condition 2 is not satisfied.

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