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3 years ago

6 1/4 % written in decimal form please

2 answers:

3 0

To find decimal form, you first have to change 1/4 into a decimal. That equals 0.25. The percentage would be 6.25%. Now to find the decimal equivalent, move the decimal two places to the left. To fill in the one missing place, add a zero before the 6. Therefore, 6 1/4% is equal to 0.0625 in decimal form.

Blizzard [7]3 years ago

3 0

First convert 6 and 1/4 to number form
6=6
1/4 divided out=0.25
6.25
percent means parts out of 100 so 6.25%=6.25/100=0.625/10=0.0625/1=0.0625

the decimal form is 0.0625

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Please help me i don’t know if I’m right

Scilla [17]

Answer:

your choice is correct

Step-by-step explanation:

The magnitude of the difference between the target amount and the actual amount will be no greater than the allowed error.

|x -34| ≤ 0.25

5 0

2 years ago

HOW DO I FIND QR<br><br> PLEASE HELP

Paul [167]

Solution: For finding QR we need to apply Pythagoras Theorem

What is Pythagoras Theorem ?

ans : Pythagoras Theorem is the sum of square of two sides which is equal to the third side or the hypotenuse. This formula is valid oy incase of Right-Angled Traingle because one of three angles here is 90°

According to this law let's apply it in the diagram shown here.

(Hypotenuse)² = (Adjacent)² + (Opposite)²

(PR)² = (QR)² + (PQ)²

(5x - 2)² = (QR)² + (3x - 1)²

(QR)² = (5x - 2)² - (3x - 1)²

After factorising both of them we get

(QR)² = 25x² - 20x + 4 - (9x² - 6x + 1)

(QR)² = 25x² - 20x + 4 - 9x² + 6x - 1

(QR)² = 16x² - 14x + 3

QR = √(16x² - 14x + 3)

So, QR is √(16x² - 14x + 3)

8 0

2 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

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

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

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

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

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

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

2 years ago

Match each equation to its solution.

motikmotik

Ik one of them.. x = 6 to 000

5 0

3 years ago

Read 2 more answers

A company provides snack-sized packages of nuts for its office employees.

DaniilM [7]

The probabilities that correctly complete this probability distribution for 50 packages of nuts will be 0.24, 0.2, 0.12, 0.28, and 0.16.

<h3>How to calculate the probability?</h3>

From the information given, we have to find the probability for each fruit. This will be:

Almond = 12/50 = 0.24

Cashew = 10/50 = 0.2

Mixed = 6/50 = 0.12

Peanut = 14/50 = 0.28

Pecan = 8/50 = 0.16

Therefore, the probabilities that correctly complete this probability distribution for 50 packages of nuts will be 0.24, 0.2, 0.12, 0.28, and 0.16.

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

6 0

1 year ago