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3 years ago

What is the inverse of the function f(x)=2x+1

Mathematics

2 answers:

Lostsunrise [7]3 years ago

7 0

Answer:The inverse of the linear function f(x)=2x+1 is f^(-1) (x) = (1/2)x-1/2

Step-by-step explanation:

liubo4ka [24]3 years ago

4 0

The answer is f^-1(x) =x/2 -1/2
The work is shown below

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For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

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3 years ago

How many millilitres in a centimeter

marta [7]

Answer:1 millilitres

Step-by-step explanation:

One millilitres equals one centimeters

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3 years ago

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vekshin1

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3 years ago

1. A researcher tested the diastolic blood pressure of 15 marathon runners and 15 non-runners. The mean for the runners was 75.9

dlinn [17]

Answer:

Calculated value t =0.0109 < t = 1.701 at 28 degrees of freedom at 0.05 level of significance.

running a good way to lower a person's blood pressure

Step-by-step explanation:

Step :1

A researcher tested the diastolic blood pressure of 15 marathon runners and 15 non-runners

The first sample size is n₁ =15

The second sample size is n₂ =15

Given data the mean for the runners was 75.9 mm Hg with an SS of 1,500

The mean of the first sample x₁⁻ = 75.9 mm

The mean of the second sample x₂⁻ = 80.3mm

The standard deviation of the first sample (S₁) = 1,500

The standard deviation of the second sample (S₂) = 8

Step 2:-

Null hypothesis :- H₀ : x₁⁻ = x₂⁻

Alternative hypothesis:- H₁ : x₁⁻ ≠ x₂⁻

level of significance :- α= 0.05

The test statistic $t = \frac{x_{1}^- -x_{2}^- }{S\sqrt{\frac{1}{n_{1} } +\frac{1}{n_{2} } } }$

where $S^{2} =\frac{n_{1}S_{1} ^2+n_{2}S_{2} ^2 }{n_{1}+n_{2}-2}$

n₁ =15 ,n₂ =15 x₁⁻ = 75.9 mm ,x₂⁻ = 80.3mm and (S₁) = 1,500 and (S₂) = 8

substitute all values in above equation, we get

$S^{2} =\frac{15X(1500) ^2+15X(8) ^2 }{15+15-2}$

s^2 = 1,205,391.42

Standard deviation = √1,205,391.42 = 1097.903

Step 3:-

The test statistic

$t = \frac{x_{1}^- -x_{2}^- }{S\sqrt{\frac{1}{n_{1} } +\frac{1}{n_{2} } } }$

x₁⁻ = 75.9 mm ,x₂⁻ = 80.3mm, n₁ =15 ,n₂ =15 and S = 1097.903

The test statistic value t = -0.01097

modulus t = 0.0109

Calculated value t =0.0109

The degrees of freedom γ=n₁+n₂ -2 = 15+15 -2 =28

From t- distribution table

From tabulated value t = 1.701 at 28 degrees of freedom at 0.05 level of significance.

Calculated value t =0.0109 < t = 1.701 at 28 degrees of freedom at 0.05 level of significance.

Therefore we accepted null hypothesis.

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