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Novosadov [1.4K]
3 years ago
5

A candy company claims that 25% of the candies in its bags are colored green. Steve buys 32 bags of 32 candies, randomly selects

one candy from each, and counts the number of green candies. If there are 7, 8, or 9 green candies, Steve will conclude that the company's claim is correct. What is the probability of Steve agreeing with the company's claim? Use Excel to find the probability.
Mathematics
1 answer:
Luden [163]3 years ago
4 0

Answer:

45.88%

Step-by-step explanation:

We have in this case a binomial distribution when p = 0.25, since the probability is 25%

. The formula that we will use in excel is BINOM.DIST, separated by commas we must put the value of x, n, p and also if it is true or false. We know that n equals 32.

In this case we need to know when x = 7, 8 and 9 and add this, and thus we will obtain the probability:

P (X = 7) + P (X = 8) + P (X = 9)

Excel formulas: = BINOM.DIST (7,32,0.25, FALSE) + BINOM.DIST (8,32,0.25, FALSE) + BINOM.DIST (9,32,0.25, FALSE)

P = 0.1546 + 0.1610 + 0.1431

P = 0.4588

Probability that steve agrees with the company's claim 45.88%

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Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

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  • ( b ) = 54 cm

  • ( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

  • <u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

\sf{   \pmb { \longrightarrow \:  \sqrt{s(s - a)(s - b)(s - c)} }}

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\begin {aligned}\quad & \quad \longmapsto  \sf  s =  \dfrac{a + b + c}{2}  \\  & \quad \longmapsto  \sf s =  \dfrac{35 + 54 + 61}{2}  \\ & \quad \longmapsto  \sf s =  \dfrac{150}{2}  \\ & \quad \longmapsto  \sf s  = 75cm \end{aligned}

Now, Area of triangle will be:

\begin{aligned}&:\implies \sf\quad \sf \:  A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \:  A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)}   \\&:\implies \sf\quad \sf \:  A = \sqrt{75 \times 40 \times 21 \times 14}  \\ &:\implies \sf\quad \sf \:  A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5}  \\ &:\implies \sf\quad \sf \:  A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5}  \\ &:\implies \sf\quad \sf \:  A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf   A =939.15 {cm}^{2} }} \end{aligned}

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

\begin{aligned}&:\implies \sf\quad \sf \:  Area =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2}  \times b \times h =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2} \times 61 \times h  = 939.15 \\&:\implies \sf\quad \sf \:  h =939.15 \times  \dfrac{2}{61}   \\&:\implies \sf\quad \sf \:  h = \dfrac{1818.3}{61}  \\ &:\implies \sf\quad  \boxed{ \pmb{\sf \:  h =30.79 \: (approx)}} \end{aligned}

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