25 multiplied by 3 algebraic expression

One kilogram is 2.2 pounds. If i had 1 pound would that be 0.45 kilograms why or why not?

prisoha [69]

Step-by-step explanation:

1 kg = 2.2 pounds

0.45 kg = 1 pounds to see whether it's correct or not we can cross multiply the given equations

multiply 1 kg with 1 pounds and 0.45 kg with 2.2 pounds then check if they are equal

1 × 1 = 2.2 × 0.45

1 = 0.99 as you can see this is not an equality therefore the statement is wrong.

3 0

4 years ago

How much higher is 500 than -400 m

gogolik [260]

Answer:

900m

Step-by-step explanation:

All you have to do is find the difference between the two measurements

500 - (-400)

Then, two negatives create a positive:

500 + 400

= 900

3 0

3 years ago

Read 2 more answers

Please someone help me out. I beggggg

bezimeni [28]

Answer:

1. 144 2. 16 3. 1 4. 3x-6

Step-by-step explanation:

So think of this as a function in a function. So you work from the inside to the outside. So for problem 1, we start with f(4)) [you read it "f of 4"] so what is the solution when x = 4, since f(x) means the function of x so f(4) means 'the function of 4' inside f(x).

Since f(x) = 3x then f(4) = 3(4) [notice how you substitute the 4 everywhere you see a letter x]

so f(4) = 12, now you work the next part h(f(4)) since f(4)=12 then h(12)

So take the h(x) function which is h(x) = $x^{2}$ then h(12) = $12^{2}$ so h(12) = 144

4 0

3 years ago

Jon has y stickers. He uses 20 stickers and divided the remainder equally between his two sisters. find the number of stickers e

ale4655 [162]

Jon has y stickers and uses 20
jon=y-20
the sisters share y-20 between 2 of them
therefore each sister receives

(y-20) divided by 2 -- (y-20)/2

6 0

3 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

3 years ago