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stepladder [879]
3 years ago
9

Do you choose to be in the honors program in middle school?

Mathematics
2 answers:
GuDViN [60]3 years ago
7 0

Answer:

You'll need to have great test scores to qualify for admission.

expeople1 [14]3 years ago
5 0

Answer:

The Honors Program is a learning contract between teachers, students, and parents. Students will take on more rigorous learning assignments within the regular classroom.

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Brad's Medicare levy was $1312.50. This was 1.5% of his taxable income.
Arisa [49]

Step-by-step explanation:

$1312.50×1.5%÷100

=20.25 that's taxable income

or

you can say $1312.50- 1.5%

=20.25

3 0
3 years ago
The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 chips
Sholpan [36]

Answer:

Using z score formula:

X = z ∂ + µ

 = 157.833

Step-by-step explanation:

Solution:

Mean = µ = 1262

Standard deviation = ∂ = 117

(a) 28th percentile for the number of chocolate chip.

P( z < z) = 28%

             = 0.28

P( z<- 0.58)  = 0.28

Z = -0.58

By using z score formula:

Z = x - µ /∂

-0.58= x – 117 / 1262

X = (- 0.58)(117) + (1262)

 =  1194.14

(b) Middle 97% of bag.

P(-z < z < z) = 97%

                    = 0.97

P( z < z) – p(z < -z) = 0.97

2p(z < z) -1 = 0.97

2p (z < z) = 1 + 0.97

P(z < z) = 1.97 / 2

            = 0.99

P(z < 2.33) = 0.99

Z ± 2.33

By using z score formula:

Z = x - µ / ∂

X = z ∂ + µ

  = - 2.33 x 117 + 1262

  =989.39

Z = 2.33

X = z ∂ + µ

  =  2.33 x 117 + 1262

 =1533.61

(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate.

By using standard normal table,

The z dist’n formula:

P(z < z ) = 25%

            =0.25

P(z < -0.6745) = 0.25

Z = 0.6745

Using z score formula:

X = z∂ + µ

   = - 0.6745 x 117 + 1262

  = 1183.0835

First quartile = Q1 =1183.0835

The third quartile is:

P(z<z) = 75%

    = 0.75

P(z < 0.6745) = 0.75

Z = 0.6745

Using z score formula:

X = z ∂ + µ

  = 0.6745 x 117 + 1262

= 1340.9165

IQR = Q3 – Q1

     = 1340.9165 – 1183.0835

  = 157.833

3 0
3 years ago
Suppose the number of radios in a household has a binomial distribution with parameters n=11 and p=40%. Find the probability of
Ivenika [448]

Answer:

Step-by-step explanation:

The formula for binomial distribution is expressed as

P(x=r) = nCr × q^(n-r) × p^r

From the information given,

n = 11

p = 40% = 40/100 = 0.4

q = 1 - 0.4 = 0.6

x represent the number of radios

a) P( x = 1) or P(x = 9)

P(x = 1) = 11C1 × 0.6^(11-1) × 0.4^1

P(x = 1) = 0.027

P(x = 9) = 11C9 × 0.6^(11-9) × 0.4^9

P(x = 9) = 0.0052

P( x = 1) or P(x = 9) = 0.027 + 0.0052 = 0.0322

b) P(x lesser than or equal to 7) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)

P(x = 0) = 11C0 × 0.6^(11-0) × 0.4^0 = 0.004

P(x = 1) = 0.027

P(x = 2) = 11C2 × 0.6^(11-2) × 0.4^2 = 0.089

P(x = 3) = 11C3 × 0.6^(11-3) × 0.4^3 = 0.177

P(x = 4) = 11C4 × 0.6^(11-4) × 0.4^4 = 0.24

P(x = 5) = 11C5 × 0.6^(11-5) × 0.4^5 = 0.22

P(x = 6) = 11C6 × 0.6^(11-6) × 0.4^6 = 0.15

P(x = 7) = 11C7 × 0.6^(11-7) × 0.4^7 = 0.15

P(x lesser than or equal to 7) = 0.004 + 0.027 + 0.089 + 0.177 + 0.24 + 0.22 + 0.15 + 0.07 = 0.977

c) P(x greater than or equal to 5) = 1 - P(x lesser than or equal to 4) = 1 - (0.004 + 0.027 + 0.089 + 0.177 + 0.24) = 1 - 0.537 = 0.463

d) P(x lesser than 9) = P(x lesser than or equal to 7) + P(x = 8)

P(x = 8) = 11C8 × 0.6^(11-8) × 0.4^8 = 0.02

P(x lesser than 9) = 0.977 + 0.02 = 0.997

e. P(x greater than 7) = 1 - P(x lesser than or equal to 7) = 1 - 0.977 = 0.023

5 0
3 years ago
the crowd at a sporting event is estimated to be 2,500 people. the exact attendance is 2,486 people. what is the percentage erro
Verdich [7]

Answer:

0.56%

Step-by-step explanation:

Percentage error

=  \frac{2500 - 2486}{2500}  \times 100 \\  \\  =  \frac{1400}{2500}  \\  \\  = 0.56\%

5 0
4 years ago
Read 2 more answers
Help me please, it would be highly appreciated! Please also explain how you got your answer.
GalinKa [24]

Answer:

answer number 2

Step-by-step explanation:

8 0
3 years ago
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