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3 years ago

8

If you a 1hour long test and you have 10 question home many minutes would you spend on each question if you used the whole hour

*15 points*

1 answer:

lord [1]3 years ago

5 0

Answer:

It would take you 10 minutes on each question. Answer is 10 minutes

Step-by-step explanation:

Because you have one hour to answer 10 questions and it should take you 10 minutes.

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Please help! Identify the recursive formula for the sequence 20, 28, 36, 44, . . . .

r-ruslan [8.4K]

Option A

Answered by GauthMath if you like please click thanks and comment thanks

8 0

3 years ago

Can't figure it out!!

Zanzabum

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)= \cfrac{1}{x} \qquad 
\begin{cases}
x_1=2\\
x_2=b
\end{cases}\implies \cfrac{f(b)-f(2)}{b-2}=-\cfrac{1}{10}
\\\\\\
\cfrac{\frac{1}{b}-\frac{1}{2}}{b-2}=-\cfrac{1}{10}\implies 
\cfrac{\frac{2-b}{2b}}{b-2}=-\cfrac{1}{10}\implies \cfrac{2-b}{2b}=-\cfrac{b-2}{10}
\\\\\\
$

$\bf \cfrac{2-b}{2b}=\cfrac{2-b}{10}\implies 20-10b=4b-2b^2\impliedby cross-multiplying
\\\\\\
2b^2-4b-10b+20=0\implies 2b^2-14b+20=0
\\\\\\
b^2-7b+10=0\implies 
(b-2)(b-5)=0\implies b=
\begin{cases}
\boxed{5}\\
2
\end{cases}$

8 0

3 years ago

I cannot understand this can you guys help

igor_vitrenko [27]

The mean is not the same thing as median
mean=78+81+85+87=331
mean=331/4=82.75
median is the middle number, but you have a par set of numbers, so your median will be the middle numbers 78,81,85,87 , which are 81+85=166, 166/2=83
83 is your median

5 0

3 years ago

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil

Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

$H_0$ : The true average penetration is 50 mils

$H_a$ : The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the <em>level of confidence</em> is α = 0.05 then the $z_\alpha$ score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)

The z-score associated with this test is

$z=\frac{\bar x-\mu}{s/\sqrt{n}}$

where

$\bar x$ = <em>mean of the sample</em>

$\mu$ = <em>average established by the specification</em>

s = <em>standard deviation of the sample</em>

n = <em>size of the sample</em>

Computing this value of z we get z = 3.42

Since z > $z_\alpha$ we can conclude that the sample has not met the required specification.

5 0

4 years ago

(2x+3y+4z)(4x^2+9y^2+16z^2-6xy-12yz-8zx)

erma4kov [3.2K]

1.1 Factoring: 4x2+9y2+16z2-6xy-12yz-8xz

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -6xy-12yz
Group 2: 16z2-8xz
Group 3: 4x2+9y2

Pull out from each group separately :

Group 1: (x+2z) • (-6y)
Group 2: (x-2z) • (-8z)
Group 3: (4x2+9y2) • (1)

Looking for common sub-expressions :

Group 1: (x+2z) • (-6y)
Group 3: (4x2+9y2) • (1)
Group 2: (x-2z) • (-8z)

6 0

3 years ago