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Nadya [2.5K]
3 years ago
14

£99 is shared in the ratio 6 : 5 What is the larger share?

Mathematics
1 answer:
nordsb [41]3 years ago
4 0

Answer:

54

Step-by-step explanation:

6x + 5x = 99

11x = 99

x=99/11

x=9

6x = 6x9 = 54

5x = 5x9 = 45

<em>ans: </em><u><em>54</em></u>

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Suppose a2 + 4a + 3 represents the area of a rectangle. What is the perimeter of that rectangle? A) 2a + 4 B) 4a + 8 C) 6a + 12
allsm [11]
A² + 4a + 3
(a+3)(a+1)         This is the length and the width of the rectangle.

Perimeter = length + length + width + width
P = (a+3) + (a+3) + (a+1) + (a+1)
P = a + a + a + a + 3 + 3 + 1 + 1
P = 4a + 6 + 2
P = 4a + 8

B) 4a + 8 
5 0
4 years ago
Please help I really need to give this in tomorrow I will give you a thanks please
zepelin [54]
A) 0+(5×2)-12+3=1
b) 0-(3×2)-1+6+7=6
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4 0
3 years ago
Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting
slega [8]

Answer:

(a) <em>λ</em> = 2.

(b) P (X = 0) = 0.1353; P (X = 1) = 0.2706;

    P (X = 2) = 0.2706; P (X = 3) = 0.1804

(c) P (Delay Problems) = 0.1431.

Step-by-step explanation:

Let <em>X</em> = number of arrivals at the drive-up teller window.

The average number of arrivals at the drive-up teller window per minute is,

<em>p</em> = 0.4 customers/ minute.

(1)

Compute the expected number of customers at the drive-up teller window in <em>n</em> = 5 minutes as follows:

E(X)=\lambda\\=np\\=5\times 0.4\\=2

Thus, the mean number of customers that will arrive in a five-minute period is <em>λ</em> = 2.

(2)

The random variable <em>X</em> follows a Poisson distribution with parameter λ = 2.

The probability mass function of <em>X</em> is:

P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...

Compute the probability of exactly 0 arrivals in 5 minutes as follows:

P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.1353\times 1}{1}=0.1353

Compute the probability of exactly 1 arrivals in 5 minutes as follows:

P(X=1)=\frac{e^{-2}2^{1}}{1!}=\frac{0.1353\times 2}{1}=0.2706

Compute the probability of exactly 2 arrivals in 5 minutes as follows:

P(X=2)=\frac{e^{-2}2^{2}}{2!}=\frac{0.1353\times 4}{2}=0.2706

Compute the probability of exactly 3 arrivals in 5 minutes as follows:

P(X=3)=\frac{e^{-2}2^{3}}{3!}=\frac{0.1353\times 8}{6}=0.1804

Thus, the values are:

P (X = 0) = 0.1353

P (X = 1) = 0.2706

P (X = 2) = 0.2706

P (X = 3) = 0.1804

(3)

Delays occur in the service time if there are more than three customers arrive during any five-minute period.

Compute the probability that there are more than 3 customers as follows:

P (X > 3) = 1 - P (X ≤ 3)

              =1-\sum\limits^{3}_{x=0}{\frac{e^{-2}2^{x}}{x!}}\\=1-(0.1353+0.2706+0.2706+0.1804)\\=1-0.8569\\=0.1431

Thus, the probability that delays will occur is 0.1431.

3 0
4 years ago
The highest possible z-score that is still in the bottom 95% of the systolic blood pressure distribution is z =−1.645 . Use this
lawyer [7]

Answer:

(1) The value of <em>X</em> is, 151.13.

This score X, is the <u>95th</u> percentile of systolic blood pressure scores among women. The percentile rank of this score is <u>95</u>.

(2) The percentage of women have systolic blood pressure in the range 140 to 159 is 9.01%.

(3) The percentage of women having systolic blood pressure below 90 is 21.19%.

Step-by-step explanation:

The random variable <em>X</em> is defined as the systolic blood pressure of women.

The distribution of <em>X</em> is, N (110, 25²).

(1)

The value of <em>z</em>-score that is in the bottom 95% is, 1.645.

Compute the value of <em>X</em> as follows:

z=\frac{X-\mu}{\sigma}\\1.645=\frac{X-110}{25}\\X=110+(1.645\times 25)\\X=151.125\\\approx151.13

The value of <em>X</em> is, 151.13.

This score X, is the <u>95th</u> percentile of systolic blood pressure scores among women. The percentile rank of this score is <u>95</u>.

(2)

Compute the probability of <em>X</em> between 140 and 159 as follows:

P(140

The percentage of systolic blood pressure between 140 and 159 is,

0.0901 × 100 = 9.01%

Thus, the percentage of women have systolic blood pressure in the range 140 to 159 is 9.01%.

(3)

Compute the probability of <em>X</em> below 90 as follows:

P(X

The percentage of women having systolic blood pressure below 90 is,

0.2119 × 100 = 21.19%.

Thus, the percentage of women having systolic blood pressure below 90 is 21.19%.

6 0
4 years ago
Find the greatest common factor.<br> 2y^3, 8y
Ira Lisetskai [31]

Answer:

STEP1:Equation at the end of step 1

2y3 - 8y

STEP2:

STEP3:Pulling out like terms

 3.1     Pull out like factors :

   2y3 - 8y  =   2y • (y2 - 4) 

Trying to factor as a Difference of Squares:

 3.2      Factoring:  y2 - 4 

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

         A2 - AB + BA - B2 =

         A2 - AB + AB - B2 =

         A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 4 is the square of 2

Check :  y2  is the square of  y1 

Factorization is :       (y + 2)  •  (y - 2) 

Final result :

2y • (y + 2) • (y - 2)

4 0
3 years ago
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