What is the value of f(5.3) f(x)=6[x] f(5.3)=

How many milliliters are in 62 liters

lesya [120]

there are 62000 millilitres in 62 liters

5 0

3 years ago

Read 2 more answers

Answer please I’m really struggling with this

katovenus [111]

To write the any polynomial in standard form you must first put it in terms separated only by $+,-$ operators, then you must order terms $x^n$ descendingly according to $n$ . In your case $-\frac{1}{5}x^2$ will be the first term because it has highest power, followed by $-8x^1$ and $+7x^0$ .

The standard form of this specific polynomial is:

$-\dfrac{1}{5}x^2-2x+7$ .

Hope this helps.

6 0

3 years ago

Rewrite without absolute value <br><br> y=|x−5|+|x+5|, if x<−5<br><br> thank you

ioda

y = | x -5| + |x+5|

as range of x is

-5 < x < 5

we can see if we put x = 0 we get

y = 10

we put any value of x in given range first term is always negative without absolute. and 2nd term is always positive

so

y = | x -5| + |x+5|

= -( x-5) + x + 5

= - x + 5 + x + 5

= 10

so

y = 10 where -5 < x < 5

8 0

4 years ago

-31.5 divided by -5 just one answer

IRINA_888 [86]

Answer:

6.3

Step-by-step explanation:

4 0

3 years ago

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A student was asked to find a 90% confidence interval for the proportion of students who take notes using data from a random sam

Scorpion4ik [409]

Answer:

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

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".

Description in words of the parameter p

$p$ represent the real population proportion of students who take notes

$\hat p$ represent the estimated proportion of students who take notes

n is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

Confidence interval

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

6 0

3 years ago