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3 years ago

Divide. Write your answer in simplest form. 1/9÷5/4=

Mathematics

2 answers:

irina [24]3 years ago

3 0

Answer:

4/45

Step-by-step explanation:

when dividing fractions, you multiply the first fraction by the second fraction's reciprocal :

1/9 divided by 5/4

1/9 multiplied by 4/5

essentially, you switch the signs from division to multiplication, and reverse the second fraction.

1/9 multiplied by 4/5

1/9*4/5 = 4/45

4vir4ik [10]3 years ago

3 0

Answer:

4/45

Step-by-step explanation:

1/9 divided by 5/4

Multiply 4 and 1

Then multiply 9 and 5

Final Answer: 4/45

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Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Evaluate a−b+3c when .a=4, b=−2, and c=3. a−b+3c =4−( )+3∙ =4+ + =

igomit [66]

Answer:

Step-by-step explanation:

4-(-2)+3(3)

4+2+9

5 0

3 years ago

Read 2 more answers

Find the least common multiple of x2 – 8x + 12 and x2 – x – 2.

abruzzese [7]

The least common multiple of x² – 8x + 12 and x² – x – 2.

by factoring:
∴ x² – 8x + 12 = (x-2)(x-6)

x² – x – 2 = (x-2)(x+1)

note: the factor (x-2) is common between them take it one time

∴ LCM = (x-2)(x-6)(x+1)

6 0

3 years ago

Read 2 more answers

a family has five children. the probability of having a girl is 1/2. whats probability of having at leasr 4 girls g'

Dima020 [189]

Answer:

Probability of having at least 4 Girls

= 0.6875

Step-by-step explanation:

Probability of having at least 4 Girls is 1-probability of having exactly 3 girls

Total number of children= 5 = N

Probability of having a girl p = 0.5

Probability of not having a girl q= 0.5

X= 3

Probability of at least 4 girls is given by

Probability= NCX(p)^x(q)^(N-x)

Probability = 5C3(0.5)^3(0.5)^(5-3)

Probability = 5C3(0.5)^3(0.5)^2

Probability= 5!/3!2!(0.5)^3(0.5)^2

Probability= 10(0.125)(0.25)

Probability= 0.3125

Probability of having at least 4 Girls

= 1- 0.3125

= 0.6875

7 0

3 years ago

Please help!!...............................

Airida [17]

B. you have to be ready to multiply all the numbers to get your answer.

3 0

3 years ago