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2 years ago

Evaluate 6+3b if b=7

2 answers:

6 0

27. Replace b with 7, making 6 + 3(7). The 3 and parentheses of 7 hint multiplication, so you multiply to make 21, then add 6 relating back to PEMDAS to teach you the correct order to solve the problem.

Wittaler [7]2 years ago

5 0

Evaluate 6+3b if b=7
To solve this problem, we must plug the value of b into the equation and follow the order of operations like so:
Original Equation: 6+3b
Plug in Given Information: 6+3*7
Remember that in the order of operations, we multiply before adding.
Multiply: 6+21
Add: The answer is 27.

You might be interested in

Find the area of the polygon WXYZ with its vertices at W(–3, –2), X(–3, 5), Y(2, 5), and Z(2, –2).

ki77a [65]

Answer:

The area of the polynomial is 35 square unit.

Step-by-step explanation:

It is given that area of the polygon WXYZ with its vertices at W(–3, –2), X(–3, 5), Y(2, 5), and Z(2, –2).

From the below graph it is noticed that the given polynomial is a rectangle with length 5 and width 7.

The distance formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the above formula we get,

$WX=7$

$XY=5$

$YZ=7$

$WZ=5$

The length of opposing sides are equal.

Using the Pythagoras theorem, the length of diagonal WY and XZ are

$WY=\sqrt{(WX)^2+(XY)^2}=\sqrt{7^2+5^2}=\sqrt{74}$

$XZ=\sqrt{(XY)^2+(YZ)^2}=\sqrt{5^2+7^2}=\sqrt{74}$

Since the length of the diagonal are same, therefore the given polygon is a rectangle.

The area of a rectangle is

$A=l\times w$

Where l is length and w is width.

The area of given polynomial is

$A=5\times 7$

$A=35$

Therefore the area of the polynomial is 35 square unit.

6 0

2 years ago

Assume there are 365 days in a year.

MissTica

Answer:

1) The probability that ten students in a class have different birthdays is 0.883.

2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.

Step-by-step explanation:

Given : Assume there are 365 days in a year.

To find : 1) What is the probability that ten students in a class have different birthdays?

2) What is the probability that among ten students in a class, at least two of them share a birthday?

Solution :

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

Total outcome = 365

1) Probability that ten students in a class have different birthdays is

The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...

$\frac{364}{365}\times \frac{363}{365} \times \frac{362}{365} \times \frac{361}{365}\times\frac{360}{365} \times \frac{359}{365} \times \frac{358}{365} \times \frac{357}{365} \times\frac{356}{365}=0.883$