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2 years ago

Find y when x = 14, if y = -3 when x = 6.

Mathematics

1 answer:

ladessa [460]2 years ago

7 0

Answer:

y=5

Step-by-step explanation:

6-(-3)=9

so...

14-9=5

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What is the answer to this AND WHY! Thank you! (−1+√5)/2

postnew [5]

Answer: .618033989 or 618033989/1000000000 or 61% (Not rounded)

Step-by-step explanation:

-1+ (the sign)5 = 1.236067977 and if you divide that by two you get .618033989 or 618033989/1000000000 or 61% (Not rounded)

6 0

3 years ago

Read 2 more answers

Place is highest to lowest 3/4 84% 1/3 0.82

sesenic [268]

ANSWER

84%,0.82,¾,⅓

EXPLANATION

Convert everything to decimals.

$\frac{3}{4} = 0.75$

$84\% = 0.84$

$\frac{1}{3} = 0.333....$

The last one is already in decimals.

$0.82$

We can now see clearly, that

$0.84 \: > \: 0.82 \: > \: 0.75 \: > \: 0.3333...$

This implies that,

$84\% \: > \: 0.82 \: > \: \frac{3}{4} \: > \: \frac{1}{3}$

Hence from highest to least, we have

84%,0.82,¾,⅓

4 0

3 years ago

In a sample of n = 6 scores, 5 of the scores are each above the mean by one point. Where is the 6th score located relative to th

den301095 [7]

The mean of a dataset is the sum of all data elements divided by the count of the elements.

The location of the 6th score relative to the mean is 5 points below the mean

Let:

$\bar x \to$ Mean

$a \to$ 5 scores

$b \to$ 6th scores

Given that:

$n = 6$

The 5 scores that are 1 above the mean implies that:

$a = \bar x + 1$

The mean of a dataset is calculated using:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x =\frac{5a + b}{6}$

$\bar x =\frac{5(\bar x + 1) + b}{6}$

Open brackets

$\bar x =\frac{5\bar x + 5 + b}{6}$

Multiply both sides by 6

$6\bar x =5\bar x + 5 + b$

Make b the subject

$b = 6\bar x -5\bar x - 5$

$b = \bar x - 5$

This means that the 6th score is 5 points below the mean

Read more about mean at:

brainly.com/question/17060266

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2 years ago

2ln(5x)=8 solve for x

White raven [17]

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3 years ago

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What fraction of box does each student receive?

Iteru [2.4K]

2.B that the answer

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3 years ago

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