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Solnce55 [7]
3 years ago
11

A multiple-choice test consists of a series of questions, each with four possible answers.

Mathematics
1 answer:
Ilya [14]3 years ago
8 0

Answer:

0.00001 ; 92

Step-by-step explanation:

Given that:

Number of possible answers / options per question = 4 ; correct answers per question = 1

P(choosing the correct answer) ; p = 1/4 = 0.25

Probability of getting atleast 30 questions right :

Mean, m = np = 60 * 0.25 = 15

Standard deviation, s = sqrt(np(1-p)) = sqrt(60 * 0.25 * 0.75) = 3.354

Using binomial approximation :

P(x ≥ 29.5) = (29.5 - 15) / 3.354 = 4.32

P(Z ≥ 4.32) = 0.00001 (Z probability calculator)

To get no more than 35%

P(x ≤ 0.35) ; normal approximation P(x ≤ 0.35n + 0.5)

m= n * 1/4 = n/4

Variance = np(1-p) = n * 1/4 * 3/4 = 3n/16

X ~(n/4, 3n/16)

P(x ≤ 0.35) = [(0.35n + 0.5 - 0.25n) / sqrt(0.1875n)] = Zcritical 99%

Zcritical at 99% = 2.326

0.1n + 0.5 / sqrt(0.1875n) = 2.326

0.1n + 0.5 = 2.326 * sqrt(0.1875n)

Square both sides :

(0.1n + 0.5)² = (2.326*sqrt0.1875n)²

Quadratic relation obtained :

0.01n² - 0.914427 + 0.25n = 0

Solving using the quadratic. Formula :

-b ± [sqrt(b² - 4ac) / 2a]

a = 0.01 ; b = - 0.914427 ; c = 0.25

Output = 91.17 or 0.27

Hence, n = 92

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arlik [135]
For question A, the answer is 2.  Since the sale sweatshirt if 40% LESS, this means that 60% of the original price still remains. 28.80 x .60 = 17.28

For question B, the answer if 4.  You can divide 17 by 20 getting .85
7 0
3 years ago
In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is
VladimirAG [237]

Answer:

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Test Statistics z = 2.65

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

P- value = 0.004025

Step-by-step explanation:

Given that:

Mean \overline x = 960 hours

Sample size n = 36

Mean population \mu = 937

Standard deviation \sigma = 52

Given that the mean  time between failures is 937 hours. The objective is to determine if the mean time between failures is greater than 937 hours

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Degree of freedom = n-1

Degree of freedom = 36-1

Degree of freedom = 35

The level of significance ∝ = 0.01

SInce the degree of freedom is 35 and the level of significance ∝ = 0.01;

from t-table t(0.99,35), the critical value = 2.438

The test statistics is :

Z = \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}}

Z = \dfrac{960-937 }{\dfrac{52}{\sqrt{36}}}

Z = \dfrac{23}{8.66}

Z = 2.65

The decision rule is to reject null hypothesis   if  test statistics is greater than  critical value.

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

The P-value can be calculated as follows:

find P(z < - 2.65) from normal distribution tables

= 1 - P (z ≤ 2.65)

= 1 - 0.995975     (using the Excel Function: =NORMDIST(z))

= 0.004025

6 0
3 years ago
Determine f(-a) where f(x) = x2 - 6x and simplify.
jeka57 [31]

Answer:

6a

Step-by-step explanation:

replace x with -a

-2a -6(-a)

-2a +8a

6a

5 0
2 years ago
7th grade math help me pleasee :))
77julia77 [94]

Answer:

step 1 communative property

Step-by-step explanation:

just look it up I forget about this

5 0
2 years ago
I need help right now!!!
antiseptic1488 [7]

Answer:

I believe it would be 4x-8=-3x+13

3 0
3 years ago
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