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3 years ago

20 divide ( 4 + 1) x 2

Mathematics

2 answers:

DIA [1.3K]3 years ago

6 0

Answer:

$\bf\huge{8}$

Step-by-step explanation:

<h2>Given that:</h2>

20 divide ( 4 + 1) x 2

simplify

<h2>formulas used:</h2>

BODMAS

where,

B = bracket
O = of
D = division
M = multiplication
A = addition
S = subtraction

<h2>explanation:</h2>

⟼ 20 ÷ ( 4 + 1 ) × 2

⟼ 20 ÷ (5) × 2

⟼ 20 ÷ 5 × 2

⟼ 4 × 2

⟼ 8

∴ 20 ÷ ( 4 + 1 ) × 2 = 8.

<h2>Hence simplified ☑️</h2>

guajiro [1.7K]3 years ago

4 0

4+1= 5
20 divide 5 x 2
5 x 2 = 10
20 divided by 10 is 2
So your answer is 2

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(a) The value of <em>k</em> is $\frac{1}{15}$ .

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

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> 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)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of <em>k</em> is $\frac{1}{15}$ .

(b)

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:

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

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

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

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

<u>Check condition 1:</u>

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

<u>Check condition 2:</u>

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

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