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Neporo4naja [7]
3 years ago
12

1/ A submarine dives to 560 meters below water level, it then dives a further 750 metres to take more deep sea pictures. What is

the total dive depth?
​
Mathematics
1 answer:
bija089 [108]3 years ago
5 0
Simply add -560 and -750 (remember same sides add and keep, different signs subtract) you should get -1310
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20 POINTS - 1 EASY QUESTION
Vaselesa [24]

Answer and explanation:

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Stephanie had seven add an 8 pound of bird seed she use three out of 8 pound to fill a bird feeder f Stephanie had 7 pound of bi
natulia [17]
So if I understand correctly, she had 7 1/8 pounds of birdseed and used 3/8 pound.
Convert the 7 1/8 to an improper fraction [(8*7)+1]/8 =57/8
57/8 - 3/8 = 54/8, which is 6 6/8 which reduces to 6 3/4 pounds left

If she had 7 pounds of birdseed and used 3/8 pounds, you need to find a common denominator to subtract if you can't do it in your head. 8 would be the common denominator, so:
56/8 - 3/8 = 53/8, which is 6 5/8 pounds left
6 0
4 years ago
Which idea was popularized in the United States by William Blackstone?
Monica [59]

Answer: Those accused of crimes should be considered innocent until

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Step-by-step explanation:

A

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3 years ago
In the figure below, Line k is perpendicular to Line l. Based on the information given in the figure, find the measure of ∠B
Serggg [28]

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Step-by-step explanation:

5 0
3 years ago
The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =
gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is \mathbf{P(X>1100n) = 0.158655}

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy  is given as :

f_x(x)  = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0

Thus, the expected  total claim amount \mu =  1000

The variance of the total claim amount \sigma ^2  = 1000^2

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})

P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})

P(X>1100n) = P(Z> \dfrac{10*100}{1000})

P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345

\mathbf{P(X>1100n) = 0.158655}

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is \mathbf{P(X>1100n) = 0.158655}

4 0
3 years ago
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