Evaluating Expressions

Mathematics

1 answer:

Svetach [21]3 years ago

6 0

1) 6
2) 2
3) -1
4) 0

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Please solve. Will mark brainilest

lorasvet [3.4K]

Answer:

$x = -3\frac{11}{20}$

Step-by-step explanation:

Convert the mixed fractions to improper fractions;

⇒ $\frac{-19}{4} = x - \frac{6}{5}$

⇒ $\frac{-71}{20} = x$ [add 6/5 to both sides]

⇒ $x = -3\frac{11}{20}$ [convert to mixed fraction]

4 0

2 years ago

Math Question: Pls help I will do anything

kow [346]

Answer:

The probability of spinning red is $\frac{1}{3}$ , the probability of spinning a 1 is $\frac{1}{6}$ , the probability of spinning an odd number is $\frac{1}{2}$ , and the probability of spinning a 9 is 0.

Step-by-step explanation:

Since there are a total of 6 possible outcomes, you can use this in the following questions. There are 2 red tiles, so the probability of spinning red is $\frac{2}{6}$ or $\frac{1}{3}$ . Since there are only one 1 tile, the probability of spinning a 1 is $\frac{1}{6}$ . Since there are three odd numbers, the probability of spinning an odd number is $\frac{1}{2}$ . Since there are no 9 in the spinner, the probability of spinning a 9 is 0 or never.

5 0

3 years ago

Find the indicated probability.

natka813 [3]

Answer:

C) 0.179

Step-by-step explanation:

Since the trials are independent, this is a binomial distribution:

Recall:

Binomial Distribution --> $P(k)={n\choose k}p^kq^{n-k}$
$P(k)$ denotes the probability of $k$ successes in $n$ independent trials
$p^k$ denotes the probability of success on each of $k$ trials
$q^{n-k}$ denotes the probability of failure on the remaining $n-k$ trials
${n\choose k}=\frac{n!}{(n-k)!k!}$ denotes all possible ways to choose $k$ things out of $n$ things