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Elis [28]
2 years ago
6

A ship captain needs to contact a diver. The communication system has a maximum range of 60 meters. Will the captain be able to

reach the diver at their current locations?
Mathematics
1 answer:
Flauer [41]2 years ago
3 0

Answer:

Step-by-step explanation:

Yes they will, although the distance given is incredulously low, but it would work nonetheless. Also, worthy of mention is that the distance is 60 m and this means that a gigantic ship would not be able to get that close to the ship, as a result of it size. And thus, only a smaller ship would be able to get that close. At a range of 60 m, the driver will surely be able to reach other ships for whatever reasons they need.

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Which of the following expressions has the greatest value when x=5? show how you arrived at your choice.
Nitella [24]

Answer:

2x^2+7

Step-by-step explanation:

Let x=5

2x^2+7

2 ( 5)^2 +7 = 2*25 +7 = 50+7 = 57

(x^3-5)/3

(5^3 -5)/3 = (125-5)/3 = 120/3 = 40

(10x -2)/ (x-3)

(10*5-2)/(5-3) = (50-2)/(2) = 48/2 = 24

6 0
3 years ago
HELP. How long is the minor axis for the ellipse shown below?
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6 0
3 years ago
If a jar of salad dressing costs $3.99 for 12 Ounces. How much does it cost per tablespoon?
romanna [79]
I ounce is 2 tablespoons so do 12 times 2 and that is 24 then multiply 3.99 times 2 thats your answer 7.98
4 0
3 years ago
Which set of directions correctly describes how to plot the point (5, 7) on the coordinate plane?
sweet [91]

Answer:3

Step-by-step explanation:

8 0
2 years ago
Box of 15 gadgets is known to contain 5 defective gadgets if 4 gadgets are drawn at random what is the probability of finding no
baherus [9]

To solve this problem, we make use of the Binomial Probability equation which is mathematically expressed as:

P = [n! / r! (n – r)!] p^r * q^(n – r)

where,

n = the total number of gadgets = 4

r = number of samples = 1 and 2 (since not more than 2)

p = probability of success of getting a defective gadget

q = probability of failure = 1 – p

 

Calculating for p:

p = 5 / 15 = 0.33

So,

q = 1 – 0.33 = 0.67

 

Calculating for P when r = 1:

P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3

P (r = 1) = 0.3970

 

 

Calculating for P when r = 2:

P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2

P (r = 2) = 0.2933

 

Therefore the total probability of not getting more than 2 defective gadgets is:

P = 0.3970 + 0.2933

P = 0.6903

 

Hence there is a 0.6903 chance or 69.03% probability of not getting more than 2 defective gadgets.

5 0
3 years ago
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