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3 years ago

Nth Term help please

Mathematics

1 answer:

liberstina [14]3 years ago

8 0

$\bf \begin{array}{|ll|ll} \cline{1-2} term&value\\ \cline{1-2} a_1&3\\[0.8em] a_2&\stackrel{3\cdot 2}{6}\\[0.8em] a_3&\stackrel{6\cdot 2}{12}\\[0.8em] a_4&\stackrel{12\cdot 2}{24}\\[0.8em] a_5&\stackrel{24\cdot 2}{48}\\[0.8em] a_6&\stackrel{48\cdot 2}{96} \\ \cline{1-2} \end{array}$

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a seed company says that 4 out of every 5 of its seeds will grow.if Madi plants 40 of these seed, then how many seeds is Madi ex

Delicious77 [7]

Answer:

38 seeds will grow

Step-by-step explanation:

4/5 = 80%

80% = .8

40*0.8 = 32

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3 years ago

Solve -4y - 3 + 3y = 8 - 2y -15, interpret the result

Elenna [48]

Answer:

y = -4

Step-by-step explanation:

-4y - 3 + 3y = 8 - 2y - 15

~Combine like terms

-y - 3 = -2y - 7

~Add 3 to both sides

-y = -2y - 4

~Add 2y to both sides

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Best of Luck!

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3 years ago

If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be

babymother [125]

Answer:

If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.

Step-by-step explanation:

Here in this question, we want to state what will happen if the null hypothesis is true in a chi-square test.

If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.

This is because at a higher level of discrepancies, there will be a strong evidence against the null. This means that it will be rare to find discrepancies if null was true.

In the question however, since the null is true, the discrepancies we will be expecting will thus be small and common.

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3 years ago

Please help! URGENT! will award 30 points & brainliest!

Alekssandra [29.7K]

Answer:

i believe it is -21, but i'm not sure

Step-by-step explanation:

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2 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago