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kramer
3 years ago
6

4) Pas sock drawer contains only black socks and white socks. There are 28 pairs of socks altogether. The ratio of bsocks towhit

e socks is 2 to 5. How many pairs of black socks does Pa own?pairs

Mathematics
1 answer:
Gekata [30.6K]3 years ago
6 0
There are 8 black socks and 20 white socks. 5 socks times 4 = 20 white socks, and 2 socks times 4 = 8 black socks. So, Pa has 8 black socks.
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I need help ASAP.
Bingel [31]

9/15

8.3%

Step-by-step explanation:

the total amount of trials was fifteen and head was scored 9 times so 9/15

the odds of a dice roll is 1/6 or 16.66% and a coin is 50%. To get both to happen in an ideal circumstance we multiply the odds so we multiply 1/6 and 1/2

8 0
3 years ago
What's 29.4983 rounded to the nearest cent
konstantin123 [22]
We need to identify what "the nearest cent" is. 

So! 
AB.CD

That is a representation of a number using variables, but we'll just say it's for place values. 

A is in the Tens Place
B is in the Ones Place
C is in the Tenths Place (1/10)
D is in the Hundredths Place (1/100

Since we are talking about money let's put it in relation to a dollar. 

A is in the Ten Dollar Place
B is in the One Dollar Place
C is in the Tenth of a Dollar Place (1/10 of a dollar)
D is in the Hundredth of a Dollar Place (1/100 of a dollar)

So, what is 1/10 of a dollar? 
What amount of money times 10, would get you 1 dollar. Or you can think of it as if you had 10 of one value of money and you got a dollar what is that? A dime. 

Now, what is 1/100 of a dollar? 
What amount of money times 100, would get you 1 dollar. 1 cent (Or it is sometimes called a penny). 

So that means any number beyond the 1/100 of a dollar point (D) will be rounded. If it's the first number after the 1/100 of a dollar is greater than (or equal to) 5 then we round the cent value up. If it is less than 5 we round down. 

$29.4983
So, 9 is our cent place. 8 is greater than 5, so we round 9 up. (Add 1. Since it is 9 it will carry over into the 1/10 of a dollar place)

Our answer is: 
$29.50

8 0
3 years ago
normally distributed with mean 24.6 seconds and standard deviation 0.64 seconds;the lower finishing time, the better. If the qua
marshall27 [118]

Answer:

z=1.04

And if we solve for a we got

a=24.6 +1.04*0.64=25.27

So the value of height that separates the bottom 85% of data from the top 15% is 25.27.  

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the finishing time of a population, and for this case we know the distribution for X is given by:

X \sim N(24.6,0.64)  

Where \mu=24.6 and \sigma=0.64

Solution to the problem

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.15   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.04. On this case P(Z<1.04)=0.85 and P(z>1.04)=0.15

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.04

And if we solve for a we got

a=24.6 +1.04*0.64=25.27

So the value of height that separates the bottom 85% of data from the top 15% is 25.27.  

3 0
3 years ago
b is the midpoint of AC. a has coordinates (9.11), and B has coordinates (-10,5). find the coordinates of c.
Vedmedyk [2.9K]
C=(-10;5). If you want an explanation I would be happy to explain it
3 0
3 years ago
Compare. Write &lt;,&gt;, or =. (Example 1)
Vitek1552 [10]

Answer:ed44

Step-by-step explanation:

4 0
3 years ago
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