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3 years ago

Need help with this question please!

Mathematics

1 answer:

nevsk [136]3 years ago

6 0

I would say it is A because if you subtract <em>p,</em> the original price by $2.50, you would get <em>d, </em>the discounted price. Look at B u see that you're adding the discount which doesn't make sense. Looking at C, the discounted price of different prices can't always be the same. And finally, D, the discounted price is greater than the original. Also, if you subtract you would get different discounts.

Read more on proportion;

brainly.com/question/1496357

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4 0

2 years ago

Mary earns $5 per hour as a waitress. Last week she took home her regular earnings of $5 per hour plus $186 in tips for a total

Iteru [2.4K]

291-186=105 (total earned through wages)

105/5=21

Therefore she worked 21 hours

6 0

3 years ago

Read 2 more answers

Cliff Branch bought a home with a 10.5% adjustable rate mortgage for 30 years. He paid $9.99 monthly per thousand on his origina

vichka [17]

I'm Unsure, But Here's What I Found.

9.99×65=649.35

10.68×65=694.2

((10.68÷9.99)−1)×100=6.9%

4 0

3 years ago

Read 2 more answers

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

What is 1 3/5 × 2 1/7

MakcuM [25]

Answer:

$\large\boxed{1\dfrac{3}{5}\times2\dfrac{1}{7}=3\dfrac{3}{7}}$

Step-by-step explanation:

Step 1:

Convert the mixed numbers to the improper fractions:

$1\dfrac{3}{5}=\dfrac{1\cdot5+3}{5}=\dfrac{8}{5}\\\\2\dfrac{1}{7}=\dfrac{2\cdot7+1}{7}=\dfrac{15}{7}$

Step 2:

We multiply the numbers remembering about simplifying:

$1\dfrac{3}{5}\times2\dfrac{1}{7}=\dfrac{8}{5\!\!\!\!\diagup_1}\times\dfrac{15\!\!\!\!\!\diagup^3}{7}=\dfrac{8\times3}{1\times7}=\dfrac{24}{7}=\dfrac{21+3}{7}=\dfrac{21}{7}+\dfrac{3}{7}=3\dfrac{3}{7}$

7 0

3 years ago