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Eddi Din [679]
3 years ago
10

Helpppp meeee! pleaseeee!​

Mathematics
2 answers:
Verdich [7]3 years ago
7 0

Answer:

2×2+5=9

2×3+5=11

2×6+5=17

2×10+5=25

garik1379 [7]3 years ago
5 0

Answer:

9

11

17

25

Step-by-step explanation:

Input the x into the equation

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Hunter-Best [27]

Answer: The correct answer is “this is a reflection across the x-axis”

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3 years ago
Y-6=-4(x+4) show steps.
Flauer [41]

Answer:

y = -4x - 10

Step-by-step explanation:

y-6=-4(x+4)

1: distributive property

y-6= -4x - 16

2: add 6 both sides

y = -4x - 10

8 0
2 years ago
Will give 5 stars on your answer and a thank you.
DENIUS [597]

Answer:

the variable term is (t)

the answer is 13.50t

4 0
2 years ago
Read 2 more answers
Determine the 1-unit growth factor for f f. [We encourage you to write the expression that represents the value of the growth fa
timofeeve [1]

Answer: The complete question is found in the attachment

Step-by-step explanation:

a) To determine the 1-unit growth factor for <em>f</em>

= 2.1/1.75 = 1.2

b) To determine the 1-unit percent change for <em>f</em>

<em>=100 x 1-1.2/1 = 0.2 x 100= 20%</em>

<em>c) to determine the initial value of f</em>

<em>f(x)= abˣ , b =1.2   a=initial value</em>

<em>at (-1,2.1)</em>

<em>2.1= (1.2)⁻¹ x a</em>

<em>a = 2.1/(1.2)⁻¹ </em>

<em>a = 2.1 x 1.2 = 2.52</em>

<em />

<em>d) to determine the function formula for f</em>

<em>f(x)= abˣ = 2.52(1.2)ˣ</em>

6 0
3 years ago
A hypothesis will be used to test that a population mean equals 10 against the alternative that the population mean is more than
PilotLPTM [1.2K]

Answer:

A) 2.58

B) 1.96

C) 1.65

Step-by-step explanation:

A hypothesis used to test that

H_o : \mu  = 10

against the alternatives H1 : \mu not equal to 10

if null hypothesis is true, then distribution of test statics follow

Zo = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}

for two sided alternatives  hypothesis ( H1: \mu \neq 7), then P value is

P = 2[1- \phi(\left | Zc \right |)]                          (1)

a)significance level  \alpha = 0.01

from 1 eq we get

\pi (\left | Zc \right |) = P(Zo

Therefore  \left | Zc \right | = 2.58    FROM Z TABLE

B) significance level  \alpha = 0.05

from 1st equation we get

\pi (\left | Zc \right |) = P(Zo < \left | Zc \right |) = 1 - \frac{0.05}{2}  = 0.975

Therefore  \left | Zc \right | = 1.96    FROM Z TABLE

C) significance level  \alpha = 0.10

from 1 eq we get

\pi (\left | Zc \right |) = P(Zo

Therefore  \left | Zc \right | = 1.65    FROM Z TABLE

3 0
3 years ago
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