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3 years ago

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At a DBE lecture of 100 students, there are 29 women and 23 men. Out of these students, 4 are teachers and 24 are either men or

teachers. Find the number of women teachers attending the lecture
.

1 answer:

sukhopar [10]3 years ago

7 0

Answer:

100-29=71

71-23=48

48/2=24

Therefore, 24-4=20

answer =20

Step-by-step explanation:

twenty teachers are the one attending the the lectures

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Psychologist Michael Cunningham conducted a survey of university women to see whether, upon graduation, they would prefer to mar

bagirrra123 [75]

Answer:

A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\$

A. P(x=6)

$P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\$

B. P(x≥10)

$P(x\geq10)=1-P(x$

$P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\$

$P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\$

$P(x\geq10)=1-(0.0008+0.0063+0.0251+0.0641+0.1175+0.1645+0.1828+0.1654+0.1241+0.0781)\\\\P(x\geq10)=1-0.9287=0.0713$

C. The expected value (mean) and standard deviation of this binomial distribution can be calculated as:

$E(x)=\mu=n\cdot p=25\cdot 0.25=6.25\\\\\sigma=\sqrt{np(1-p)}=\sqrt{25\cdot 0.25\cdot 0.75}=\sqrt{4.69}\approx2.17$

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2 years ago

HELP ASAP PLEASE <br> Not understanding and need answers asap

Neko [114]

Answer:

you domb if you dont know the answer B

Step-by-step explanation:

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2 years ago

O Watch help video

Ostrovityanka [42]

Answer:

325+250= 575

Step-by-step explanation:

4 0

2 years ago

11.1 x 0.8 x 1.16 x 0.4

makkiz [27]

Answer: 4.12032 which is also equivalent to 4:12

3 0

3 years ago

1. Average Senior High School annual cost of tuition fee for all private schools in Cavite last year was 46,300 a random sample

anygoal [31]

Answer:

Step-by-step explanation:

H0 : μ = 46300

H1 : μ > 46300

α = 0.05

df = n - 1 = 45 - 1 = 44

Critical value for a one tailed t-test(since population standard deviation is not given).

Tcritical = 1.30

The test statistic :(xbar - μ) ÷ (s/sqrt(n))

The test statistic, t= (47800-46300) ÷ (5600√45)

t = 1500

t = 1500 / 834.79871

t = 1.797

The decision region :

Reject H0: if t value > critical value

1. 797 > 1.30

Tvalue > critical value ; We reject H0

Hence, there is sufficient evidence to conclude that cost has increased.

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2 years ago