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Diano4ka-milaya [45]
2 years ago
8

Once a delay or disability is diagnosed, the best thing to do is to continue to be a family’s knowledgeable and reliable partner

in child care.Once a delay or disability is diagnosed, the best thing to do is to continue to be a family’s knowledgeable and reliable partner in child care.
Mathematics
1 answer:
meriva2 years ago
5 0

Once you detect that there is a delay or disability, then you should continue to be the reliable partner in child care to the family so this is True.

<h3>What should be done when a delay in child development is seen?</h3>

The observation and screening process can lead to a child care partner discovering a delay or disability.

When this happens, you should report to your supervisor to check if the child is eligible for federal and state programs related to their condition. Whatever the case, you should remain accessible to the family as their partner in child care.

Find out more on child development delays at brainly.com/question/5082527

#SPJ1

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The dimensions of a wading pool shaped like a rectangular prism are shown below. What is the volume of the pool?
beks73 [17]

Answer:

126

Step-by-step explanation:

l*w*h

l=6

w=5

h=3.5

3.5*5=21

21*6=126

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9 apples there are 6 fewer apples than oranges
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If your trying to find the amount of oranges there would be 15
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. A shipyard makes a container ship that can withstand the total amount of weight W, which is normally distributed with mean of
Debora [2.8K]

Answer:

the maximum number of containers that the ship can load is 170

Step-by-step explanation:

Given the data in the question;

W ~ N( 600, 60 )

S ~ N( 4n, 0.4√n )

so

p( W > S ) = 0.90

⇒ P( W - S > 0) = 0.9 ------ let this be equation 1

now, since W and S are independent

Mean( W - S ) = Mean( W ) - Mean( S 0 = 600 - 4n

and

SD( W - S ) = √( var(W) + var(S) ) = √( 60² + 0.4²n)

hence;

W - S ~ N( 600 - 4n, √( 60² + 0.4²n) )

now, from equation one, P( W - S > 0) = 0.9

P( \frac{(W-S)-Mean(W-S)}{SD(W-S)} > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90

P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = 0.90

from z- table

P( z > \frac{0-(600-4n)}{\sqrt{60^2 + 0.4^2n} }) = P( z >-1.282)  

\frac{4n - 600}{\sqrt{60^2 + 0.4^2n} } = -1.282 ------------------ let this be equation 2

now, we square both sides of equation 2

\frac{(4n - 600)^2}{60^2 + 0.4^2n} } = (-1.282)^2  

\frac{(4n - 600)(4n-600)}{3600 + 0.16n} } = 1.643524

we cross multiply

16n² + 360000 - 4800n = 1.643524( 3600 + 0.16n )

16n² + 360000 - 4800n = 5916.6864 + 0.26296384n

16n² + 360000 - 5916.6864 - 4800n - 0.26296384n = 0

16n² + 354083.3136 - 4800.26296384n = 0    

16n² - 4800.26296384n + 354083.3136 = 0  

solving the quadratic equation, we know that;

x = -b±√( b² - 4ac ) / 2a

so we substitute

x = [-(-4800.26296384) ±√( (-4800.26296384)² - (4 × 16 × 354083.3136)] / [2×16]

x = [ 4800.26296384 ±√( 23042524.522 - 22661332.0704 ] / 32

x = [ 4800.26296384 ±√(381192.4516) ] / 32

x = [ 4800.26296384 ± 617.4078 ] / 32

Hence;

x = [ 4800.26296384 - 617.4078 ] / 32 or  [ 4800.26296384 + 617.4078 ] / 32

x = 131   or  170

Therefore, the maximum number of containers that the ship can load is 170

5 0
3 years ago
12)
Rina8888 [55]

Answer:

-0.55-0.05

Step-by-step explanation:

When you solve (0.25-0.3)-(0.8-0.25), you get -0.6. -0.55-0.05= -0.6

8 0
4 years ago
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