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Ksju [112]
3 years ago
8

1/2x + 1/4 = 6 PLEASE help

Mathematics
1 answer:
NemiM [27]3 years ago
5 0

Answer:

11.5 or 11 1/2

Step-by-step explanation:

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If 9 x 0 + 4 = 4 had parenthesis where would they go?
Bingel [31]

Answer:

(9 x 0) + 4=4

Step-by-step explanation:

5 0
3 years ago
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A partial proof was constructed given that MNOP is a parallelogram.
Elena-2011 [213]

Answer:

B

Step-by-step explanation:

B. angle N is supplementary to angle P

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3 years ago
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A cookie recipe requires 3 teaspoons of baking soda for 36 cookies. If the baker would like to make 480 cookies, how much baking
Vitek1552 [10]

Answer:

It would require 40 teaspoons

Step-by-step explanation:

Given that;

A cookie recipe requires 3 teaspoons of baking soda for 36 cookies

The amount of teaspoons of baking soda required per cookie is;

r = 3/36 teaspoons per cookie

So, for 480 cookies i would require;

N = r × 480 cookies

Substituting r, we have;

N = 3/36 teaspoons per cookie × 480 cookies

N = 40 teaspoons

It would require 40 teaspoons

4 0
3 years ago
Anna is draining water out of a 50 gallon rain barrel. The graph shows the relationship between time in minutes and the gallons
tatiyna

Answer:

6 < x ≤ 20

Step-by-step explanation:

First, let's use the points (0, 50) and (20, 0) to find the equation of the line.

m = Δy / Δx

m = (0 − 50) / (20 − 0)

m = -2.5

y = -2.5x + 50

Now find the time when y = 35.

35 = -2.5x + 50

-15 = -2.5x

x = 6

So there is less than 35 gallons in the barrel when 6 < x ≤ 20.

7 0
3 years ago
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
3 years ago
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