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3 years ago

What percent of 78 = 39?

Mathematics

1 answer:

S_A_V [24]3 years ago

3 0

39/78 = .5 = 50%
...................

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Find all numbers such that three times the number is greater than eight less than the number.

klio [65]

Answer:

ALL SUCH NUMBERS WHICH ARE GREATER THAN -4 satisfy the given condition.

Step-by-step explanation:

Here, assume such number = m

Now, given :

3 times the number = 3 x ( m) = 3 m

8 less than the given number = m - 8

Now, according to the question:

3 times the number > 8 less than the given number

or, 3 m > m - 8

or, 3m - m > m - 8 + m

or, 2 m > - 8

or, m > - 4

Hence, all SUCH NUMBERS WHICH ARE GREATER THAN -4 satisfy the given condition.

6 0

3 years ago

Last month Ed ate 9 apples five bananas 4 peaches and 7 oranges find the Ratio of bananas to the total number of fruit then expl

Soloha48 [4]

The ratio is 5:4:7:9. The reason why they didn't write it together is that the bananas are written in different form.

8 0

3 years ago

A cake that costs $32 is on sale for 15% off and there are no taxes. Explain why the expression

Darya [45]

Answer:

$27.2

Step-by-step explanation:

A cake that costs $32 is on sale for 15% off and there are no taxes. Explain why the expression (0.85).32 represents the final price of the cake in dollars.

This is explained in the calculation below

= A cake that costs $32 is on sale for 15% off

The amount off the cake is calculated as

= 15% of $32

= 15/100 × 32

= $4.8

The amount he is paying for the cake =>$32 - $4.8

= $27.2

The above expression is obtained as

(0.85).32

= (100% - 15%) × 32

=( 1 - 0.15) × 32

= 0.85 × 32

= $27.2

The final price of the cake = $27.2

7 0

3 years ago

Indicate the data relationship for each table. If each is, directly proportional or non-proportional

Alla [95]

It’s nonporportiniol sorry if it’s spelled wrong lol

5 0

3 years ago

Suppose that a box contains 8 cameras and that 4 of them are defective. A sample of 2 cameras is selected at random with replace

Dafna1 [17]

The Expected value of XX is 1.00.

Given that a box contains 8 cameras and that 4 of them are defective and 2 cameras is selected at random with replacement.

The probability distribution of the hypergeometric is as follows:

$P(x,N,n,M)=\frac{\left(\begin{array}{l}M\\ x\end{array}\right)\left(\begin{array}{l}N-M\\ n-x\end{array}\right)}{\left(\begin{array}{l} N\\ n\end{array}\right)}$

Where x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

The probability distribution for X is obtained as below:

From the given information, let X be a random variable, that denotes the number of defective cameras following hypergeometric distribution.

Here, M = 4, n=2 and N=8

The probability distribution of X is obtained below:

The probability distribution of X is,

$P(X=x)=\frac{\left(\begin{array}{l}5\\ x\end{array}\right)\left(\begin{array}{l}8-5\\ 2-x\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}$

The probability distribution of X when X=0 is

$\begin{aligned}P(X=0)&=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}8-4\\ 2-0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-0)!0!}\right)\times \left(\frac{4!}{(4-2)!2!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end$

The probability distribution of X when X=1 is

$\begin{aligned}P(X=1)&=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}8-4\\ 2-1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-1)!1!}\right)\times \left(\frac{4!}{(4-1)!1!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.57\end$

The probability distribution of X when X=2 is

$\begin{aligned}P(X=2)&=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}8-4\\ 2-2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}4\\ 0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-2)!2!}\right)\times \left(\frac{4!}{(4-0)!0!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end$

Use E(X)=∑xP(x) to find the expected values of a random variable X.

The expected values of a random variable X is obtained as shown below:

The expected value of X is,

E(X)=∑xP(x-X)

E(X)=[(0×0.21)+(1×0.57)+(2×0.21)]

E(X)=[0+0.57+0.42]

E(X)=0.99≈1

Hence, the binomial probability distribution of XX when X=0 is 0.21, when X=1 is 0.57 and when X=2 is 0.21 and the expected value of XX is 1.00.

Learn about Binomial probability distribution from here brainly.com/question/10559687

#SPJ4

8 0

2 years ago