Final Answer The number of students tested in the city is 150. The probability of a student testing positive with at least two strains of malaria parasites is . The departmen t of health should not close the schools because the probability of a student testing positive with at least two strains is below 45%. Question 1 Step 1: Determine the Number of Students Tested To find the total number of students tested, we need to consider all the students who tested positive for any of the strains and those who tested negative. Let: be the set of students who tested positive for Celin parasites. be the set of students who tested positive for Euglin parasites. be the set of students who tested positive for Tetrin parasites. Given: Students who tested positive for Tetrin only = 20 Students who tested negative for any strain = 40 Using the principle of inclusion-exclusion, the total number of students who tested positive for at least one strain is: We need to find . Since 20 students tested positive for Tetrin only, we have: Substituting the known values: Now, substitute back to find : The total number of students tested is: Step 2: Calculate the Probability of Testing Positive with at Least Two Strains The number of students who tested positive with at least two strains is: The probability is: Step 3: Advise the Department of Health The probability of a student testing positive with at least two strains is , which is approximately 23.33%. Since this is below the 45% threshold, the schools should not be closed. Answer The number of students tested in the city is 150. The probability of a student testing positive with at least two strains of malaria parasites is . The department of health should not close the schools because the probability of a student testing positive with at least two strains is below 45%. No Ads and higher accuracy? Upgrade Was the solution helpful ? YES NO

Mathematics

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Step 1: Kuamua Idadi Kamili ya Wanafunzi Waliochunguzwa
Ili kupata idadi kamili ya wanafunzi waliochunguzwa, tunajumlisha wanafunzi wote walioonyesha dalili chanya kwa aina yoyote ya vimelea na wale walioonyesha dalili hasi, kulingana na mchoro wa Venn.
Idadi ya wanafunzi walioonyesha dalili chanya kwa angalau aina moja:

15(Celin pekee)+10(Euglin pekee)+20(Tetrin pekee)+5(Celin na Euglin pekee)+10(Celin na Tetrin pekee)+5(Euglin na Tetrin pekee)+5(Celin, Euglin na Tetrin)=7015 (\text{Celin pekee}) + 10 (\text{Euglin pekee}) + 20 (\text{Tetrin pekee}) + 5 (\text{Celin na Euglin pekee}) + 10 (\text{Celin na Tetrin pekee}) + 5 (\text{Euglin na Tetrin pekee}) + 5 (\text{Celin, Euglin na Tetrin}) = 70

Idadi ya wanafunzi walioonyesha dalili hasi kwa aina yoyote ni 40.
Jumla ya wanafunzi waliochunguzwa ni:

70+40=11070 + 40 = 110

Part a· 1 step

Idadi ya wanafunzi waliochunguzwa katika jiji ni 110\boxed{\text{110}}.

  1. Kuhesabu Uwezekano wa Mwanafunzi Kuwa na Angalau Aina Mbili za Vimelea vya Malaria

    Idadi ya wanafunzi walioonyesha dalili chanya kwa angalau aina mbili za vimelea vya malaria inapatikana kwa kujumlisha maeneo ya makutano ya seti mbili au tatu katika mchoro wa Venn:

    5(Celin na Euglin pekee)+10(Celin na Tetrin pekee)+5(Euglin na Tetrin pekee)+5(Celin, Euglin na Tetrin)=255 (\text{Celin na Euglin pekee}) + 10 (\text{Celin na Tetrin pekee}) + 5 (\text{Euglin na Tetrin pekee}) + 5 (\text{Celin, Euglin na Tetrin}) = 25

    Uwezekano wa mwanafunzi kuonyesha dalili chanya kwa angalau aina mbili za vimelea vya malaria ni:

    P(angalau aina mbili)=Idadi ya wanafunzi wenye angalau aina mbiliJumla ya wanafunzi waliochunguzwaP(\text{angalau aina mbili}) = \frac{\text{Idadi ya wanafunzi wenye angalau aina mbili}}{\text{Jumla ya wanafunzi waliochunguzwa}} P(angalau aina mbili)=25110P(\text{angalau aina mbili}) = \frac{25}{110}

    Kurahisisha sehemu:

    P(angalau aina mbili)=522P(\text{angalau aina mbili}) = \frac{5}{22}

Part b· 1 step

Uwezekano wa mwanafunzi kuonyesha dalili chanya kwa angalau aina mbili za vimelea vya malaria ni 522\boxed{\frac{5}{22}}.

  1. Kushauri Idara ya Afya

    Uwezekano wa mwanafunzi kuonyesha dalili chanya kwa angalau aina mbili za vimelea vya malaria ni 522\frac{5}{22}.
    Kubadilisha sehemu kuwa asilimia:

    522≈0.22727×100%≈22.73%\frac{5}{22} \approx 0.22727 \times 100\% \approx 22.73\%

    Kizingiti cha kufunga shule ni 45%45\%.
    Kwa kuwa 22.73%22.73\% iko chini ya 45%45\%, idara ya afya haipaswi kufunga shule.

    c) Idara ya afya haipaswi kufunga shule kwa sababu uwezekano wa mwanafunzi kuonyesha dalili chanya kwa angalau aina mbili za vimelea ni takriban 22.73%\text{22.73\%}, ambayo iko chini ya kizingiti cha 45%\text{45\%}.

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