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2 years ago

How many 5-digit numbers are there that are divisible by either 45 or 60 but are not divisible by 90?

Mathematics

1 answer:

o-na [289]2 years ago

3 0

Answer:

2500 numbers

Step-by-step explanation:

Problem;

Solving for the number of 5-digit numbers divisible either by 45 or 60 but not divisible by 90;

Let us approach this in a step wise manner;

5 -digits numbers ranges from 10,000 to 99,999

To solve this problem, let us find the number of digits that are divisible by 45;

the first number divisible by 45 within this range is 10035

the last number divisible by 45 in this range is 99990

increment is 45

The total number in this range is:

$\frac{last number - first number}{increment}$ + 1 = $\frac{99990 - 10035}{45}$ + 1

= 2000 numbers

To solve this problem, let us find the number of digits that are divisible by 60;

the first number divisible by 60 within this range is 10020

the last number divisible by in this range is 99960

increment is 60

$\frac{99990 - 10080 }{90}$

The total number in this range is:

$\frac{last number - first number}{increment}$ + 1 = $\frac{99960 - 10020}{60}$ + 1

= 1500 numbers

To solve this problem, let us find the number of digits that are divisible by 90;

the first number divisible by 90 within this range is 10080

the last number divisible by in this range is 99990

increment is 90

The total number in this range is:

$\frac{last number - first number}{increment}$ + 1 = $\frac{99990 - 10080}{90}$ + 1

= 1000 numbers

Now solution:

Number of 5 digits = number of 45 + number of 60 - number of 90

= 2000 + 1500 - 1000

= 2500 numbers

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Answer:

The mean number of successful surgeries is 1.57.

The variance of the number of successful surgeries is 0.3111.

Step-by-step explanation:

STEP 1

If the tear on the left knee has a rim width of less than 3mm, the probability that the surgery on the left knee will be successful (ls) is 0.90.

That isP(Is)=0.90

The probability that the surgery on the left knee will fail (lf) is 0.10. That isP(V)=0.10

If the tear on the right knee has a rim width of 3-6 mm, the probability that the surgery on the right knee will be successful (rs) is 0.67. That isP(rs)=0.67

The probability that the surgery on the right knee will fail (rf) is 0.33.

That isP(vf)= 0.33

Let the random variable X denote the number of successful surgeries. The range of X is{0,1,2)

Now find the probabilities associated with the possible values of X. The number of successful surgeries is equal to 0 if the surgeries on both knees fail. Since the surgeries are independent, we have:

P(X = 0)=P(lf and rf)

= P()P(rf)

=(0.10)(0.33)

= 0.033

The number of successful surgeries is equal to 1 if the surgery on one knee is successful and the surgery on the other knee fails, that is

P(X =1)= P((ls and rf) or (if and rs))

= P(Is)P(rf)+P(V)P(rs)

= (0.90)(0.33)+(0.10)(0.67)

= 0.364

The number of successful surgeries is equal to 2 if the surgeries on both knees are successful, that is

P(X = 2)=P(ls and rs)

= P(Is) P(rs)

=(0.90)(0.67)

= 0.603

So, the probability mass function of X is the following

X 0 1 2

f(x) 0.033 0.364 0.603

The mean number of successful surgeries is,

E(X)=∑xP(x)

=(0×0.033) +(1×0.364)+(2×0.603)

=1.57

the mean number of successful surgeries is 1.57

The expected value is obtained by taking the summation of the product of each possible value of the random variable X with its corresponding probabilities. Thus, the mean number of successful surgeries is 1.57.

STEP 2

The variance of the number of successful surgeries is,

Var(X)=∑x²P(x)-(E(x²)

=(0×0.033) + (1 × 0.364) +(2 × 0.603) - (1.57)²

= 2.776 -(1.57)²

=0.3111

The variance of the number of successful surgeries is 0.3111.

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During a field trip, 60 students are put into equal-sized groups. Describe two ways to interpret 60÷5 60 ÷ 5 in this context. Fi

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The second question:

Consider the division expression $7\frac{1}{2} / 2$ . Select all multiplication equations that correspond to this division expression.

$2 * ? = 7\frac{1}{2}$ $7\frac{1}{2} * ?= 2$ $? * 2 = 7\frac{1}{2}$

$2* 7\frac{1}{2} = ?$ $? * 7\frac{1}{2} = 2$

Answer:

1. See Explanation

2. $2 * ? = 7\frac{1}{2}$ and $? * 2 = 7\frac{1}{2}$

Step-by-step explanation:

Solving (a):

Given

$Students = 60$

$Group = Equal\ Sized$

Required

Interpret $\frac{60}{5}$ in 2 ways

Interpretation 1: Number of groups if there are 5 students in each

Interpretation 2: Number of students in each group if there are 5 groups

Solving the quotient

$Quotient = \frac{60}{5}$

$Quotient = 12$

For Interpretation 1:

The quotient means: 12 groups

For Interpretation 2:

The quotient means: 12 students

Solving (b):

Given

$7\frac{1}{2} / 2$

Required

Select all equivalent multiplication equations

Let ? be the quotient of t $7\frac{1}{2} / 2$

So, we have:

$7\frac{1}{2}/2 = ?$

Multiply through by 2

$2 * 7\frac{1}{2}/2 = ? * 2$

$7\frac{1}{2} = ? * 2$

Rewrite as:

$? * 2 = 7\frac{1}{2}$ --- This is 1 equivalent expression

Apply commutative law of addition:

$2 * ? = 7\frac{1}{2}$ --- This is another equivalent expression

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Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

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F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

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F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

For X=0

This happens when every couple and the singles are on time (ot).

$P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602$

For X=1

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

$P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908$

For X=2

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

$P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704$

For X=3

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

$P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055$

For X=4

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

$P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285$

For X=5

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

$P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550$

For X=6

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

$P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427$

For X=7

This happens when

1) Only one of the singles is on time (2 combinations)

$P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390$

For X=8

This happens when everybody is late

$P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147$

8 0

2 years ago