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3 years ago

What is the answer to 1/2-1/4

Mathematics

2 answers:

irinina [24]3 years ago

4 0

1/2-1/4
(1/2x2)-1/4
2/4-1/4
2-1
1
1/4

monitta3 years ago

4 0

Hi , the answer is 1/4 , because we have to find the common denominator which is 4 .Then we have to change the fractions , 1/2 is now 2/4 and 1/4 stays the same , then we just subtract.
1/2
2 times 2 equals 4

1 times 2 equals 2
2/4

1 times 1 equals 1
4times 1 equals 4
1/4

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X^2=1 has two solutions. If the equation is changed to 4x^2=1 which operation find the solutions of the new equation?

snow_lady [41]

Answer:

X = -1/2 or 1/2

Step-by-step explanation:

When the square root of a number is found, this usually results in 2 solutions which may be equal in size but opposite in signs.

As such,

X^2=1 to find x find the root of both sides

X = √1

X = -1 or 1

Going by this

4x^2=1

Divide both sides by 4

X^2=1 /4

taking the square root of both sides

X = √1/4

The root of 1 is -1 or 1 while that of 4 is -2 or 2 hence,

X = -1/2 or 1/2

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3 years ago

Read 2 more answers

Write the simplified expression.

Vesna [10]

Answer:

343

Step-by-step explanation:

Let's start simplifying with the numerator:

(15-8)⁹

=7⁹

=40353607

Now simplify the denominator :

[(5+2)³]²

=[(7)³]²

=[343]²

=117649

Now combine them :

(15-8)⁹/ [(5+2)³]²

=40353607 / 117649

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6 0

4 years ago

The mean weight of 10 randomly selected newborn babies at a local hospital is 7.14 lbs and the standard deviation is 0.87 lbs. A

sergejj [24]

Answer:

a) $ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$

b) 90% of all samples of size 10 have sample means within 0.5035 of the population mean.

c) The 90% confidence interval would be given by (6.636;7.643)

d) Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

e) If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=7.14$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.87 represent the sample standard deviation

n=10 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that $t_{\alpha/2}=1.83$

$ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$

Part b

90% of all samples of size 10 have sample means within 0.5035 of the population mean.

Part c

Now we have everything in order to replace into formula (1):

$7.14-1.83\frac{0.87}{\sqrt{10}}=6.637$

$7.14+1.83\frac{0.87}{\sqrt{10}}=7.643$

So on this case the 90% confidence interval would be given by (6.636;7.643)

Part d

Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

Part e

If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

7 0

4 years ago