Answer:
68%
Step-by-step explanation:
<em><u>68-95-99.7% Rule:</u></em>
<em><u /></em>
This is the empirical rule which is used to remember the percentage of values that is within a band of the mean. We say:
- 68% of the data fall within 1 standard deviation of the mean
- 95% of the data fall within 2 standard deviation of the mean, and
- 99.7% of data falls within 3 standard deviations of the mean
Clearly, from the empirical rule, we see that about 68% of the area is between z = -1 and z = 1 (or within 1 standard deviation of the mean)
Answer:
13 = x 15 - x = 2
Step-by-step explanation:
15-x=2
15-13=2
The ratio between the surfaces areas of the two prisms is equal to 6.25.
<h3>
How to get the ratio of the surface areas?</h3>
The ratio between the volumes of the prisms is:

This number is equal to the cube of the scale of dilation applied to the dimensions of the smaller prism, so the scale factor is:
![\sqrt[3]{15.625} = 2.5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B15.625%7D%20%3D%202.5)
And the ratio between the areas should be the square of the scale factor, then we will have that the ratio between the surface areas of the two prisms is:

If you want to learn more about scale factors, you can read:
brainly.com/question/3457976
All angles = 180 degrees
3(x-2)+35+52=180
distribute
3x-6+35+52=180
combine like terms
3x+81=180
subtract 81 on both sides
3x=99
divide both sides by 3
x=33
plug in for x
3((33)+2)=
or 180-35-52=93
<h2>
Answer:</h2>

<h2>
Step-by-step explanation:</h2>
As the question states,
John's brother has Galactosemia which states that his parents were both the carriers.
Therefore, the chances for the John to have the disease is = 2/3
Now,
Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.
Now, one of those children married with a person.
So,
Probability for the child to have disease will be = 1/2
Now, again the child's child (Martha) probability for having the disease is = 1/2.
Therefore,
<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)
<em><u>Hence, the probability for the first child to have Galactosemia is
</u></em>