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lilavasa [31]
3 years ago
13

What number is d in 5d-7= -13?

Mathematics
1 answer:
Andru [333]3 years ago
3 0
5d - 7 = -13
5d = -13 + 7
5d = -6
d = -6/5
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Joe wants to put a fence around his rectangular garden. his garden measures 33 ft by 39 ft. the garden has a path around it that
sukhopar [10]
The perimeter of a rectangle is given by:
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¿cuáles son los criterios de congruencia y semejanza de triángulo?
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Answer:

Two triangles are similar if they meet one of the following criteria. : Two pairs of corresponding angles are equal. : Three pairs of corresponding sides are proportional. : Two pairs of corresponding sides are proportional and the corresponding angles between them are equal.

Dos triángulos son similares si cumplen uno de los siguientes criterios. : Dos pares de ángulos correspondientes son iguales. : Tres pares de lados correspondientes son proporcionales. : Dos pares de lados correspondientes son proporcionales y los ángulos correspondientes entre ellos son iguales.

Step-by-step explanation:

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3 years ago
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In a test of the effectiveness of garlic for lowering​ cholesterol, 8181 subjects were treated with raw garlic. Cholesterol leve
xz_007 [3.2K]

Answer:

With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

Step-by-step explanation:

The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired <em>t</em>-test is used to determine the effectiveness of garlic for lowering​ cholesterol.

A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.

The hypothesis for the test can be defined as follows:

<em>H₀</em>: With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.

<em>Hₐ</em>: With garlic​ treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.

The information provided is:

\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01

Compute the test statistic value as follows:

t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22

The test statistic value is 0.22.

Decision rule:

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value of the test as follows:

p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132

*Use a <em>t</em>-table.

The <em>p</em>-value of the test is 0.4132.

<em>p-</em>value= 0.4132 > <em>α</em> = 0.01

The null hypothesis was failed to be rejected.

Thus, it can be concluded that with garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

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Answer:

D. translation, yes

Step-by-step explanation:

A is out because the shapes are congruent, and if the figure was reflected, it would have been closer to the y-axis. It is the same with B, if the figure was dilated, the second figure would have been similar but not congruent. C is out because the figure is still facing the same direction.

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For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant
nirvana33 [79]

Answer:

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of y with respect to x over the interval by definition of secant line:

r = \frac{y(b) -y(a)}{b-a} (1)

Where:

a, b - Lower and upper bounds of the interval.

y(a), y(b) - Function exaluated at lower and upper bounds of the interval.

If we know that y = 3\cdot x^{2}, a = 3 and b = 6, then the average rate of change of y with respect to x over the interval is:

r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}

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The average rate of change of y with respect to x over the interval [3,6] is 27.

b) The instantaneous rate of change can be determined by the following definition:

y' =  \lim_{h \to 0}\frac{y(x+h)-y(x)}{h} (2)

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h - Change rate.

y(x), y(x+h) - Function evaluated at x and x+h.

If we know that x = 3 and y = 3\cdot x^{2}, then the instantaneous rate of change of y with respect to x is:

y' =  \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}

y' =  3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}

y' = 3\cdot  \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}

y' = 6\cdot  \lim_{h \to 0} x +3\cdot  \lim_{h \to 0} h

y' = 6\cdot x

y' = 6\cdot (3)

y' = 18

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

5 0
2 years ago
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