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lions [1.4K]
2 years ago
14

How do I solve this problem 34=p+19

Mathematics
1 answer:
NARA [144]2 years ago
4 0

Hello,

34 = p + 19

34 - 19 = p + 19 - 19

15 = p

p = 15

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Joey completed 1,456 push-ups over a 4-week period. Joey did the same number of push-ups every day. How many push-ups did Joey d
kkurt [141]

Answer:

364

Step-by-step explanation: i divided so if its not correct you can kill me later

8 0
3 years ago
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Help with distance formula
Anna35 [415]
Length= \sqrt {(x1-x2)^2+(y1-y2)^2} \\ \\ =\sqrt {(-5-(-2))^2+(1-3)^2} \\ \\ =\sqrt {-3^2+-2^2} \\ \\ = \sqrt {13} \\ \\ = 3.61 units
5 0
4 years ago
Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi
katrin2010 [14]

Answer:

Test scores of 10.2 or lower are significantly low.

Test scores of 31.4 or higher are significantly high.

Step-by-step explanation:

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:

\mu = 20.8, \sigma = 5.3

Identify the test scores that are significantly low or significantly high.

Significantly low

Z = -2 and lower.

So the significantly low scores are thoses values that are lower or equal than X when Z = -2. So

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

-2 = \frac{X - 20.8}{5.3}

X - 20.8 = -2*5.3

X = 10.2

Test scores of 10.2 or lower are significantly low.

Significantly high

Z = 2 and higher.

So the significantly high scores are thoses values that are higherr or equal than X when Z = 2. So

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

2 = \frac{X - 20.8}{5.3}

X - 20.8 = 2*5.3

X = 31.4

Test scores of 31.4 or higher are significantly high.

3 0
3 years ago
Help please thank you
Leto [7]
Ah, this my friend, is actually easier than it looks. I promise. Sort of. XD

Alright, so let's start with the basics. You have two shapes that look congruent, and obviously ARE congruent, but how they are congruent can be different.
HIJ (Shape 1) is congruently equal to (~=) shape LKJ (The order does indeed matter) by what?

Well, in terms of congruency, you have about 8 different ways, I only remember 4.
SSS (Side, Side, Side)
SAS (Side, Angle, Side)
ASA (Angle, Side, Side)
AAA (Angle, Angle, Angle)

This means that whichever it is, each must be identified as congruent to the other. If it's SAS, you must know, for certain (not you personally, you can guess, but that's not what they want, they want you to know based on the info they give you) that there are 2 sides that are congruent, and 1 angle that are congruent. Same for all the others, just plug and play. 

In the text, this question mentions that side HJ is congruently equal to JL. This means you have 1 set of sides identified as congruent. 
The text ALSO mentions that angle H is also congruently equal to angle L. This means you now have 1 set of angles that are congruently equal.

So far, you know you have 1 congruent set of sides (S) and One congruent set of angles (A)

Now, you also can see that based on what we already know, HIJ extends to LKJ, meaning the other angle would ALSO be congruent. 
This leaves you with "ASA" (Angle, Side, Angle), meaning 2 sets of angles are congruent, and 1 set of sides.

Your answer is A

~Hope this helps!
6 0
4 years ago
Consider the inequality -5(x+7)<-10. Write an Inequality representing the solution for x.
alexdok [17]

It could be 3 and lower because if you - 7 from -10 you get -17 then your equation would be -5x< -17 than you divide -17 by -5 and get 3.4 than your equation is x < 3.4 so your answer is 3.4 or below.


7 0
3 years ago
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