5/7 x=40 a.28 b.96 c.20 d.100 pls help

How can you use rational numbers to represent real world problems?

Dahasolnce [82]

Their is a lot of ways to use rational numbers in the world like miles per gallon,number of pages per hr.

7 0

4 years ago

I need somebody HELP! Not just anybody HELP!

sweet-ann [11.9K]

Answer:

you didnt include the question-

3 0

3 years ago

Read 2 more answers

Angela is going on vacation to Mexico. She needs to buy new shirts for her trip. She buys 10 new shirts that cost her a total of

choli [55]

Angela bought 4 casual shirts, and 6 dressy shirts.

To find this, I used the equation ($18c) + ($25d) = $222, and filled in for c (casual) and d (dressy) until the outcomes were the same. Filling in for c and d, you'll get ($18 * 4) + ($25 * 6) = $222. Solving both sides, you'll get $72 + $150 = $222 or $222 = $222. Thus meaning she bought 4 casual shirts and 6 dressy shirts,

I hope this helps!

7 0

3 years ago

If a student was an undergraduate Business major, what is the probability that the student intends to attend classes full-time i

labwork [276]

Answer:

0.701

Step-by-step explanation:

Probability is the ratio of favorable outcomes to the total number of outcomes. It provides incite to likelihood an event to occur. Probability is a ratio , so it has no unit. It is sometimes expressed in percentage.

probability that the student intends to attend classes full time in pursuit of an MBA degree = 352/502 = 0.7012.

6 0

3 years ago

A recent kellogg survey shows that approximately 34% of americans eat breakfast. suppose alex takes a random sample of four amer

vlabodo [156]

This kind of experiments are ruled by Bernoulli's formula. If you have probability p of "success", and you want k successes in n trials, the probability is

$\binom{n}{k}p^k(1-p)^{n-k}$

It's easier to compute the first probability by difference: instead of computing the probability of the event "at least one of the surveyed eats breakfast", let's compute the probability of its contrary: none of them eats breakfast. So, we want 0 successes in 4 trials, with probability of success 0.34. The formula yields

$\binom{4}{0} 0.34^0 0.64^4 = 0.64^4 \approx 0.17 = 17\%$

Since the contrary has probability 17%, our event "at least one of the surveyed eats breakfast" has probability 83%.

As for the second question, the event "at least three of the surveyed eats breakfast" is the union of the events "exactly three of the surveyed eats breakfast" and "exactly four of the surveyed eats breakfast". So, we just need to sum their probabilities:

$\binom{4}{3} 0.34^3 0.64^1 +\binom{4}{4} 0.34^4 0.64^0 = 4\cdot 0.34^3 0.64 + 0.34^4 \approx 0.11 = 11\%$

3 0

3 years ago