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3 years ago

Name three solutions of x > 10.5

Mathematics

1 answer:

Svetllana [295]3 years ago

5 0

In the problem is says that d is greater than 10.5 so any value greater than 10.5 is the answer

X = 11
X= 12
X= 14.

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A-2.5<br> 2a-22.7<br> a+1.3<br> 21.5<br> What’s a

rewona [7]

Answer:

2.4

Step-by-step explanation:

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3 years ago

If y= (2x^2+x)^3 +(x-1) /(1-x). find dy dx

Mrrafil [7]

Answer:

Step-by-step explanation:

y=(2x²+x)³+(x-1)/(1-x)

=(2x²+x)³+(x-1)/-(x-1)

=(2x²+x)³-1

dy/dx=3(2x²+x)(4x+1)-0

=3x(2x+1)(4x+1)

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3 years ago

When graphed, which parabola opens downward? y = –3x2 y = (x – 3)2 y = x2 – 3

In-s [12.5K]

The answer to this question is y = –3x^2, just had this on e2020 :)

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3 years ago

Read 2 more answers

What is the simplest form of 4/7+1/3=

Serggg [28]

Answer:

19/21 or .90

Step-by-step explanation:

Common denominator.

Multiply 4/7 by 3 which equals to 12/21

Multiply 1/3 by 7 which equals to 7/21

Add 12/21 and 7/21

Answer: 19/21

3 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