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3 years ago

I really don't understand all of this

Mathematics

2 answers:

dalvyx [7]3 years ago

7 0

Answer:

$5(x + 2y - 3)$

Step-by-step explanation:

$the \: answer \: while \\ \: expanding \: will \: get \: like \: this \\ \\ 5(x + 2y - 3) \\ = (5 \times x) +( 5 \times 2y )+ (5 \times - 3) \\ = 5x + 10y - 15 \\ \\ hope \: it \: helps \\ plzzz \: mark \: it \: as \: brainliest...$

PolarNik [594]3 years ago

5 0

Answer:

Step-by-step explanation:

Distribute each one like a robot

Or factor it like a human.

5(x+2y - 3)

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Pre Calculus Help! Will give Brainliest!

Sveta_85 [38]

It's a quotient identity

6 0

3 years ago

The combined average weight of an okapi and a llama is 450450 kilograms. The average weight of 33 llamas is 190190 kilograms mor

Nataliya [291]

Answer:

Okapi 290 kg

Llama 160 kg

Step-by-step explanation:

Let weight of each llama be $l$

Let weight of each okapi be $o$

Given, combined weight of 1 okapi and 1 llama is 450, we can write:

$l+o=450$

Also, average weight of 3 llama is 190 more than the average weight of 1 okapi, thus we can write:

$3l-190=o$

Now, substituting 2nd equation into 1st equation, we can solve for weight of 1 llama:

$l+o=450\\l+(3l-190)=450\\l+3l-190=450\\4l=450+190\\4l=640\\l=\frac{640}{4}=160$

Each llama weights 160 kg, now using this and plugging into 2nd equation, we get weight of 1 okapi to be:

$o=3l-190\\o=3(160)-190\\o=290$

Each okapi weigh 290 kg

5 0

3 years ago

if Michael had four oranges and Timmy had 5 oranges and they combined them together how many oranges do they have in total

bija089 [108]

They have 9 because
$5 + 4 = 9$

4 0

2 years ago

Read 2 more answers

20% of US High School teens vape. A local High School has implemented campaigns to reduce vaping among students and believes tha

zaharov [31]

Answer:

10.93% probability of observing 51 or fewer vapers in a random sample of 300

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .