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Julli [10]
3 years ago
12

A recent study found that 81?% of? breast-cancer cases are detectable by mammogram. Suppose a random sample of 11 women with bre

ast cancer are given mammograms. Find the probability that all of the cases are? detectable, assuming that detection in the cases is independent.
Mathematics
1 answer:
elena-s [515]3 years ago
7 0

Answer:

9.847 %

Step-by-step explanation:

Be p1 the probability that the mammogram detect breast cancer in a woman.

p1=81%.

The probability 'P' that independent and consecutive 'n' singular events with p probability occur is:

P=p_{1} ^{n}

In this case:

P: Probability that breast cancer be detected on 11 women with breast cancer.

P=0.81^{11}=0.09848

P=9.847 %

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Answer:

The range of T is a subspace of W.

Step-by-step explanation:

we have T:V→W

This is a linear transformation from V to W

we are required to prove that the range of T is a subspace of W

0 is a vector in range , u and v are two vectors in range T

T = T(V) = {T(v)║v∈V}

{w∈W≡v∈V such that T(w) = V}

T(0) = T(0ⁿ)

0 is  Zero in V

0ⁿ  is zero vector in W

T(V) is not an empty subset of W

w₁, w₂   ∈ T(v)

(v₁, v₂ ∈V)

from here we have that

T(v₁) = w₁

T(v₂) = w₂

t(v₁) + t(v₂) = w₁+w₂

v₁,v₂∈V

v₁+v₂∈V

with a scalar ∝

T(∝v) = ∝T(v)

such that

T(∝v) ∈T(v)

so we have that T(v) is a subspace of W. The range of T is a subspace of W.

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Step-by-step explanation:

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