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2 years ago

What is 1 half subtract 1 tenth

2 answers:

8 0

To solve for $\frac{1}{2}- \frac{1}{10}$ You first need to find a common denominator.

To do so, you need to make both denominators 10 by multiplying the top and bottom of $\frac{1}{2}$ by 5

$\frac{5}{10}- \frac{1}{10}$ = $\frac{4}{10}$ Reduce by dividing both the top and bottom by 2

Your answer is $\frac{2}{5}$

Yuliya22 [10]2 years ago

6 0

The answer is 0.4 that is what i got

You might be interested in

Brandon uses the steps below to solve the equation 15x+6=14x+5 using algebra tiles. Step 1 Add 14 negative x-tiles to both sides

dsp73

Answer:

No he was not correct

Step-by-step explanation:

The Equation is 15 x + 6 = 14 x + 5

now By the steps brandon used let us solve the question

Step 1 : Add 14 negative x tile to both sides

which will make the equation

15x + 6 - 14x = 14x + 5 - 14 x

x + 6 = 5 .......(ii)

Step 2: Add 5 negative unit tiles to both sides of equation (ii)

which will make the equation

x +6 - 5 = 5 - 5

x + 1 = 0 .........(iii)

Step 3: x =1 is the answer

but from equation (iii) we see that

x + 1 =0

by adding -1 on both sides

x + 1 -1 = 0-1

x = -1

So the step three done by Brandon is wrong which explains he was not correct

6 0

2 years ago

Read 2 more answers

Determine the interquartile range for the given data 11, 13, 16, 17, 8, 10, 14, 14, 17, 14

Elanso [62]

Answer:

(IQR) interquartile range: 4.5!

Step-by-step explanation:

8, 10, 11, 13, 14, 14, 14, 16, 17

Median: 14

Lower quartile: 10.5

Upper quartile: 15

Interquartile range: 15 - 10.5 = 4.5

7 0

2 years ago

The mean number of words per minute (WPM) read by sixth graders is 97 with a standard deviation of 19 WPM. If 75 sixth graders a

andrey2020 [161]

Answer:

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ ;

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 97, \sigma = 19, n = 75, s = \frac{19}{\sqrt{75}} = 2.1939$

What is the probability that the sample mean would be greater than 101.63 WPM?

This is 1 subtracted by the pvalue of Z when X = 101.63. So

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

By the Central Limit Theorem

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

$Z = \frac{101.63 - 97}{2.1939}$

$Z = 2.11$

$Z = 2.11$ has a pvalue of 0.9826.

1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

5 0

2 years ago

If I can travel 280 miles in 4.5 hours how many hours will it take me to travel 500 miles round to the nearest hundred

Lelu [443]

Answer: 8.01

Step-by-step explanation:

Speed = distance/time = 280/4.5 = 62.44mph

Time = distance/speed = 500/62.44 = 8.0071 hrs

6 0

2 years ago

Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a)

Anna35 [415]

Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.

Given the function is $f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}$ and the Rolle's Theorem does not apply to the function.

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The condition for Rolle's theorem to be true as:

f(a)=f(b)
f(x) must be continuous in [a,b].
f(x) must be differentiable in (a,b).

To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.

If we look closely at the given function we can see that the first derivative of the given function is:

$\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end$

From this point of view we can see that the given function is not defined for x=0.

Hence, all the assumptions are not satisfied we can reach a conclusion that we cannot apply the Rolle's Theorem.

Learn more about Rolle's Theorem from here brainly.com/question/12279222

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8 0

1 year ago