Convert. 5 days = lao hours

What is the answer? Y= -x-4 Y=x

Mashcka [7]

Do you mean like this?

Y = -x - 4
Y = x

x = -x - 4
2x = -4
x = -2

Therefore y = -2

7 0

3 years ago

The grid shows Figure Q and its image Figure Q′ after a transformation.

Svetach [21]

The transformation of pentagon Q to pentagon Q' is a clockwise rotation of 180° about the origin.

<h3>What is transformation?</h3>

A transformation is a general term for four specific ways to manipulate the shape or position of a point, a line, or a geometric figure.

We have,

Coordinates of a pentagon Q.

(2, 4), (3, 7), (7, 5), (5,4), and (4,2).

Coordinates of pentagon Q'.

(-2, -4), (-3, -7), (-7, -5), (-5,-4), and (-4,-2).

We see that the coordinates of each point of the pentagon Q have changed their position as negative coordinates in pentagon Q'

The coordinates of pentagon Q are in (x, y) which can be assumed to be in the first quadrant of the coordinate plane.

The coordinates of pentagon Q' are in (-x, -y) which is in the 3rd quadrant.

From the first quadrant to 3rd quadrant there is a rotation of 180°.

Thus the transformation of pentagon Q to pentagon Q' is a clockwise rotation of 180° about the origin.

Learn more about transformation here:

brainly.com/question/4458799

#SPJ1

8 0

1 year ago

it is known that the population proton of utha residnet that are members of the church of jesus christ 0l6 suppose a random samp

Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Proportion of 0.6

This means that $p = 0.6$

Sample of 46

This means that $n = 46$

Mean and standard deviation:

$\mu = p = 0.6$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722$

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

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

By the Central Limit Theorem

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

$Z = \frac{0.5 - 0.6}{0.0722}$

$Z = -1.38$

$Z = -1.38$ has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

8 0

3 years ago

You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which

stellarik [79]

To model this situation, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial deposit
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$t$ is the time in years

For account A:
We know for our problem that $P=2000$ and $r= \frac{2.25}{100} =0.0225$ . Since the interest is compounded monthly, it is compounded 12 times per year; therefore, $n=12$ . Lets replace those values in our formula:
$A=2000(1+ \frac{0.0225}{12} )^{12t}$

For account B:
$P=2000$ , $r= \frac{3}{100} =0.03$ , $n=12$ . Lest replace those values in our formula:
$A=2000(1+ \frac{0.03}{12} )^{12t}$

Since we want to find the time, $t$ , when the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
 $2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000$

Now that we have our equation, we just need to solve for $t$ :
$2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000$
$(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}$
$(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}
{2}$
$ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})$
$12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})$
$t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})$
$t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}$
$17.47$

We can conclude that after 17.47 years the sum of the balance in both accounts will be at least 5000.

5 0

3 years ago

How can i factorise <img src="https://tex.z-dn.net/?f=12x%20-%206" id="TexFormula1" title="12x - 6" alt="12x - 6" align="absmidd

Lorico [155]

Answer:

6(2x - 1)

Step-by-step explanation:

Given

12x - 6 ← factor out common factor of 6 from both terms

= 6(2x - 1) ← in factored form

8 0

3 years ago

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Convert.5 days =laohours​

Convert.5 days =laohours