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serg [7]
3 years ago
14

Can I get some help on these please

Mathematics
2 answers:
Vadim26 [7]3 years ago
7 0
Here you are. 
See attachement. :)

Tema [17]3 years ago
6 0
Calculate y-values by inserting x-values.
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At Basketball practice Kyla made 60 free throws this was 80% of her attempts how many free throws did kyle attempt?
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5 0
2 years ago
Every evening Jenna empties her pockets
rodikova [14]

Answer:

15

Step-by-step explanation:

1 dime = 10

1 quarter=25

x=number of coins - dimes

y= number of coins - quarters

x (10)+y(25)=10.25

x+y=50

x= 50-y

x(10)+y(25)=1025

(50-y)(10)+25y =1025

500-10y+25y=1025

500+15y=1025

15y= 1025 - 500

15y=525

y=525/15

y=35

x= 50-35=15

3 0
2 years ago
Is (-5, 6) a solution to y = 2x + 4 ?<br> I need an answer quickly, please
Fittoniya [83]

Answer:

(-5, 6) is not a solution.

Step-by-step explanation:

6=2(-5)+4

6=-10+4

-10+4=-6, not 6.

7 0
2 years ago
Angle BAC measures 56. What is the measure of angle BDC?
goldfiish [28.3K]

Answer: Like the angles BAC (56°) and BDC has the same arc BC in the circumference, these angles must be congruent, then angle BDC must be equal to 56°.

4 0
3 years ago
Read 2 more answers
Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If​ women's weights are
Mariulka [41]

Answer:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

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

X \sim N(173.6,49.8)  

Where \mu=173.6 and \sigma=49.8

We are interested on this probability

P(141.8

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0
3 years ago
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