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2 years ago

Consider the following, what is the summation of all solutions? (2x+4)^2+1=5

Mathematics

1 answer:

Rus_ich [418]2 years ago

8 0

Answer:

- 4

Step-by-step explanation:

(2x + 4)² + 1 = 5 ← expand left side using FOIL and simplify

4x² + 16x + 16 + 1 = 5

4x² + 16x + 17 = 5 ( subtract 5 from both sides )

4x² + 16x + 12 = 0 ( divide terms by common factor 4 )

x² + 4x + 3 = 0 ← in standard form

(x + 3)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x + 1 = 0 ⇒ x = - 1

Summation of solutions = - 3 + (- 1) = - 3 - 1 = - 4

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Step-by-step explanation:

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A triangle has side lengths of (6.2p + 4.5q) meters, (3.4p – 0.7r) meters, and (2.9r – 4.1q) meters. Which expression represents

Natalija [7]

Answer:

9.6p + 0.4q + 2.2r meters (answer A)

Step-by-step explanation:

We sum up the three side lengths to obtain the perimeter of the triangle:

(6.2p + 4.5q) meters, , and (2.9r – 4.1q) meters

+ (3.4p – 0.7r) meters

+ (2.9r – 4.1q) meters

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9.6p + 0.4q + 2.2r meters (answer A)

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2 years ago

Which of the following hypothetical scenarios best describes a scenario where approximation by the binomial distribution is most

VladimirAG [237]

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2 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.

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

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3 years ago