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Novosadov [1.4K]
2 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]2 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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Craig ran the first part of a race with an average speed of 8 miles per hour and biked the second part of a race with an average
lutik1710 [3]
Let the distance of the first part of the race be x, and that of the second part, 15 - x, then
x/8 + (15 - x)/20 = 1.125
5x + 2(15 - x) = 40 x 1.125
5x + 30 - 2x = 45
3x = 45 - 30 = 15
x = 15/3 = 5

Therefore, the distance of the first part of the race is 5 miles and the time is 5/8 = 0.625 hours or 37.5 minutes
The distance of the second part of the race is 15 - 5 = 10 miles and the time is 1.125 - 0.625 = 0.5 hours or 30 minutes.
4 0
2 years ago
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A football quarterback enjoys practicing his long passes over 40 yards. He misses the first pass 40% of the time. When he misses
dangina [55]

Answer:

8% or 0.08

Step-by-step explanation:

Probability of missing the first pass = 40% = 0.40

Probability of missing the second pass = 20% = 0.20

We have to find the probability that he misses both the passes. Since the two passes are independent of each other, the probability that he misses two passes will be:

Probability of missing 1st pass x Probability of missing 2nd pass

i.e.

Probability of missing two passes in a row = 0.40 x 0.20 = 0.08 = 8%

Thus, there is 8% probability that he misses two passes in a row.

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2 years ago
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The student council at Spend too much High School is planning a school dance. The table below shows the budget for the dance.
hodyreva [135]
$300 income
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$300 - $350 = a loss of $50
6 0
2 years ago
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A sample of size 60 from one population of weights had a sample average of 10.4 lb. and a sample standard deviation of 2.7 lb. A
Olegator [25]

Answer:

z=  0.278

Step-by-step explanation:

Given data

n1= 60 ; n2 = 100

mean 1= x1`= 10.4;     mean 2= x2`= 9.7

standard deviation  1= s1= 2.7 pounds ;  standard deviation 2= s2 = 1.9 lb

We formulate our null and alternate hypothesis as

H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)

We set level of significance α= 0.05

the test statistic to be used under H0 is

z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂

the critical region is z > ± 1.96

Computations

z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100

z= 10.4- 9.7/ √ 7.29/60 + 3.61/100

z= 0.7/√ 0.1215+ 0.0361

z=0.7 /√0.1576

z= 0.7 (0.396988)

z= 0.2778= 0.278

Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0  at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.

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Write a comparison sentence that represents this equation 5x9=45
mel-nik [20]
9+9+9+9+9=45 I'm not sure if that's what they're asking for
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