Using the normal distribution relation, the probability that sample will exceed the weight limit is 0.004
<u>Using the relation</u> ::
The mean, μ = np = (162 × 19) = 3078
The standard deviation, σ = 28 × √19 = 122.049
<u>The Zscore</u> :
Zscore = (3401 - 3078) ÷ (122.049)
Zscore = 2.65
Hence,
P(Z > 2.65) = 1 - P(Z < 2.65)
Using a normal distribution table :
P(Z > 2.65) = 1 - 0.9959
P(Z > 2.65) = 0.004
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Answer:
Step-by-step explanation:
Let
c -----> the additional amount (in dollars) David will spend
we know that
The word " at most" in this context means "less than or equal to"
The amount David has spent plus possible additional amounts he will spend must be less than or equal to $39
so
The inequality that represent this situation is
solve for c
subtract 22 both sides
The maximum amount he could spend is $17
Factors of 84: 1, 2<span>, </span>3<span>, 4, 6, </span>7<span>, 12, </span>14<span>, </span>21<span>, </span>28<span>, </span>42<span>, 84. Prime factorization: 84 = </span>2<span> x </span>2<span> x </span>3<span>x </span>7<span> which can also be written (</span>2^2<span>) x </span>3<span> x </span>7<span>.</span>