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Studentka2010 [4]

1 year ago

10

Mapiya writes a series of novels. She earned \$75{,}000$75,000dollar sign, 75, comma, 000 for the first book, and her cumulative

earnings double with each sequel that she writes.

1 answer:

qwelly [4]1 year ago

4 0

Answer:

$E(n)=75000 \times 2^n$

Complete question:

write a function that gives mapiyas cumulative earnings E(n), in dollars when she has written n sequel's

Step-by-step explanation:

According to the question, she earned $75000 for the first book.

Also,We are given that her cumulative earnings double with each sequel that she writes.

Assuming she has written n sequel's

Now since we are given that her cumulative earnings double with each sequel

So, her initial earning will be $2^n$ times

So, her earning will be : $75000 \times 2^n$

Now we are given that cumulative earnings is denoted by E(n)

So, the function becomes : $E(n)=75000 \times 2^n$

Hence a function that gives Mapiya's cumulative earnings E(n), in dollars when she has written n sequel's is $E(n)=75000 \times 2^n$

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The probability of flowering plants bearing flowers in January is given the table. If a plant has flowered in January, what is t

ololo11 [35]

Answer:

Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

Step-by-step explanation:

We know that:

Probability =Total favourable outcomes/ Total possible outcomes

Probability of flowering plants bearing flowers in January is given in the table. It means that out of 100 peony 70 peony flowers in January.

There are five kinds of flowers.

We have to select peony and find it's Probability.

As all flowers are independent, their flowering phenomenon does not depend on each other.

So,Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

7 0

1 year ago

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A caravan crossed 1,378 miles of desert in 85 days. it traveled 22 miles on the first day and 28 miles on the second day. if the

vlabodo [156]

Um idk I forgot how to do this but ima ask my brother for you so he can tell you ok

8 0

2 years ago

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Of the bundles delivered daily to the store, 1/3

Triss [41]

Answer:

x=9

Step-by-step explanation:

Let total bundle=x

Morning edition of the daily sun=1/3x

=x/3

Afternoon edition of the daily sun=2

That leaves 2x/3 - 2, of which

x/3 - 1 are regional

1 is local

1 from another state

Sum everything

x = x/3 + 2 + x/3 - 1 + 1 + 1

x=2x/3 +3

x-2x/3=3

3x-2x/3=3

x/3=3

Cross product

x=3×3

x = 9

3 0

2 years ago

Marketing companies have collected data implying that teenage girls use more ring tones on their cell phones than teenage boys d

ASHA 777 [7]

Answer:

The p-value here is 0.0061, which is very small and we have evidence that the girls' mean is higher than the boys' mean.

Step-by-step explanation:

We suppose that the two samples are independent and normally distributed with equal variances. Let $\mu_{1}$ be the mean number of ring tones for girls, and $\mu_{2}$ the mean number of ring tones for boys.

We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} > 0$ (upper-tail alternative).

The test statistic is

T = $\frac{\bar{X}_{1}-\bar{X}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$

where

$S_{p} = \sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}$ .

For this case, $n_{1}=n_{2}=20$ , $\bar{x}_{1}=3.2$ , $s_{1}=1.5$ , $\bar{x}_{2}=2.2$ , $s_{2}=0.8$ .

$s_{p} = \sqrt{\frac{(19)(1.5)^{2}+(19)(0.8)^{2}}{38}} = 1.2021$ and the observed value is

t = $\frac{3.2-2.2}{1.2021\sqrt{1/20+1/20}} = \frac{1}{0.3801} = 2.6309$ .

We can compute the p-value as P(T > 2.6309) where T has a t distribution with 20 + 20 - 2 = 38 degrees of freedom, so, the p-value is 0.0061. Because the p-value is very small, we can reject the null hypothesis for instance, at the significance level of 0.05.

3 0

1 year ago

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

8 0

1 year ago

Read 2 more answers