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aleksklad [387]
3 years ago
8

On a primary school parents were asked the number of hours they spend to help their children in home work there were 90 parents

who helped 1/2 hour to 1 1/2 hours . 20% of parents helped more than 1 1/2 hours 30% said that they helped for 1/2 hour to 1 1/2 hours per day the rest 50% did not help at all
(i)how many parents were surveyed
(ii)how many said they did not help
Mathematics
2 answers:
puteri [66]3 years ago
7 0

50 parents were surveyed

20 said they did not help

plz mark brainly

plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

lilavasa [31]3 years ago
5 0

Answer:

Step-by-step explanation:

Let x be the number of parents surveyed.

30% of x = 90

\frac{30}{100}*x=90\\\\x=90*\frac{100}{30}=3*100=300\\

Total number of parents surveyed = 300

Number of parents who did not help = 50%of 300

=\frac{50}{100}*300=50*3=150

Number of parents who did not help = 150

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Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
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Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

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Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

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From the standard normal distribution tables,

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P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

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