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3 years ago

Someone pls help me solve this problem. No guessing Use the photo below to solve for x

Mathematics

1 answer:

skad [1K]3 years ago

7 0

Answer:

x = 6

Step-by-step explanation:

9/x = 12/8

12x = 72

/12 /12

x = 6

hope this helps ^^

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Salsk061 [2.6K]

Answer:

18 is divisible by both 2 and 9, you can multiply 9 by 2 to get 18, and 18 is a multiple of both of those numbers

Step-by-step explanation:

5 0

3 years ago

Second pictures of my last question

Elanso [62]

Answer:

you really did not give us the full pictuer

Step-by-step explanation:

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2 years ago

An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha

TEA [102]

Answer:

The probability is $P(X >300 ) = 0.97219$

Step-by-step explanation:

From the question we are told that

The capacity of an Airliner is k = 300 passengers

The sample size n = 320 passengers

The probability the a randomly selected passenger shows up on to the airport

$p = 0.96$

Generally the mean is mathematically represented as

$\mu = n* p$

=> $\mu = 320 * 0.96$

=> $\mu = 307.2$

Generally the standard deviation is

$\sigma = \sqrt{n * p * (1 -p ) }$

=> $\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }$

=> $\sigma =3.50$

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

$P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )$

Here $\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )$

=> $P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )$

Now applying continuity correction we have

$P(X >300 ) = P(Z > \frac{[300+0.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > -1.914 )$

From the z-table

$P(Z > -1.914 ) = 0.97219$

$P(X >300 ) = 0.97219$

8 0

3 years ago

What is the greatest common factor of 44c^5, 22c^3 and 11c^4

Alexxandr [17]

$\text{ GCF of } 44c^5, 22c^3 , 11c^4 \text{ is } 11c^3$

Solution:

Given that, we have to find the greatest common factor of $44c^5, 22c^3 , 11c^4$

The greatest number that is a factor of two (or more) other numbers.

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor

$\underline{\text{ GCF of } 44c^5, 22c^3 , 11c^4 :}$

First find the GCF of numbers and then find the GCF of variables and then multiply them together

Step 1: GCF of 44, 22 and 11

The factors of 11 are: 1, 11

The factors of 22 are: 1, 2, 11, 22

The factors of 44 are: 1, 2, 4, 11, 22, 44

11 is the greatest factor that is common in above three list

Then the greatest common factor is 11

$\text{ Step 2: GCF of }c^5, c^3 , c^4$

$c^5 = c^3 \times c^2\\\\c^3 = c^2 \times c^1 \text{ or } c^3\\\\c^4 = c^3 \times c^1$

Thus the greatest common factor is $c^3$

Step 3: Multiply the GCF of 44, 22, 11 and GCF of $c^5, c^3 , c^4$

$\text{ GCF of } 44c^5, 22c^3 , 11c^4 = 11 \times c^3 = 11c^3$

Thus $\text{ GCF of } 44c^5, 22c^3 , 11c^4 \text{ is } 11c^3$

8 0

3 years ago

Simplify (9x + 10) - (7x + 8)

Ilia_Sergeevich [38]

Answer:

2x+2

Step-by-step explanation:

of symbols, representing a value, o relation. example 2+2=4

8 0

3 years ago