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Tcecarenko [31]

3 years ago

5

PLEASE HELP ME!!!! PLEASE!!!!!!!!

2 answers:

weqwewe [10]3 years ago

8 0

Answer:

Both have the same probabilities of having a meal that includes salmon

Step-by-step explanation:

As they both ordered one main dish, one aske for one of each column, and only in the main dish there´s salmon, and the other one directly asked for one main dish, in the main dish column there´s salmon, so they both have a 1/4 probability of getting salmon.

Kamila [148]3 years ago

6 0

Sondera cause he/she could get salmon and more but grichen cant cause he olny ordered from 1

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Use the Distributie<br> Property to express<br> 27 + 60

Lunna [17]

Once you add 27+60, the answer is 87

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3 years ago

Read 2 more answers

If 25=4x-7, then x =

sasho [114]

Answer:

x=8

Step-by-step explanation:

Move all terms that don't contain x to the right side and solve.

6 0

3 years ago

I just can't do this

Roman55 [17]

You can, it's not as hard as you might think

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4 0

3 years ago

Question 1 (Essay Worth 10 points) (01.02 MC) Part A: If (26)x = 1, what is the value of x? Explain your answer. (5 points) Part

katen-ka-za [31]

Answer:

See Explanation

Step-by-step explanation:

The question is not clear. However, I will treat the question as:

$(26)x = 1$

$(50)x = 1$

and:

$(2^6)^x = 1$

$(5^0)^x = 1$

Solving: $(26)x = 1$ and $(50)x = 1$

$(26)x = 1$

Divide both sides by 26

$x = \frac{1}{26}$

$(50)x = 1$

Divide both sides by 50

$x = \frac{1}{50}$

Solving $(2^6)^x = 1$ and $(5^0)^x = 1$

$(2^6)^x = 1$

Express 1 as 2^0

$(2^6)^x = 2^0$

Remove bracket

$2^{6x} = 2^0$

Cancel out 2

$6x = 0$

Divide both sides by 6

$x = \frac{0}{6}$

$x = 0$

$(5^0)^x = 1$

Express 1 as 5^0

$(5^0)^x = 5^0$

Cancel out 5^0

$x = 1$

8 0

3 years ago

Helpppppppppppppppppppp

Sloan [31]

Answer:

Sum of $\frac{7x}{x^2-4}$ and $\frac{2}{x+2}$ is $\mathbf{\frac{9x-4}{x^2-4}}$

Option B is correct answer.

Step-by-step explanation:

We need to find sum of $\frac{7x}{x^2-4}$ and $\frac{2}{x+2}$

Finding sum of $\frac{7}{x^2-4}$ and $\frac{2}{x+2}$ :

$\frac{7x}{x^2-4}+\frac{2}{x+2}$

We know that $x^2-4 =(x+2)(x-2)$

Replacing x^2-4

$\frac{7x}{(x+2)(x-2)}+\frac{2}{x+2}$

Now, taking LCM of (x+2)(x-2) and (x+2) we get (x+2)(x-2)

$=\frac{7x+2(x-2)}{(x+2)(x-2)}\\=\frac{7x+2x-4}{(x+2)(x-2)}\\=\frac{9x-4}{(x+2)(x-2)}\\=\frac{9x-4}{x^2-4}$

So, Sum of $\frac{7x}{x^2-4}$ and $\frac{2}{x+2}$ is $\mathbf{\frac{9x-4}{x^2-4}}$

Option B is correct answer.

6 0

3 years ago