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1 month ago

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Aubree wanted to see if there is a connection between the time a given exam takes place and the average score of this exam. She

collected data about all the exams from the previous year and arranged the average scores of the exams according to the period they took place, as shown below . (PLEASE HELP .. WHOEVERS FAST AND RIGHT WILL BE BRAINLIEST)

2 answers:

Svet_ta [12.7K]1 month ago

5 0

Answer:

you can compute the average by splitting the results by the exam duration and arriving at your answer

Step-by-step explanation:

PIT_PIT [12.4K]1 month ago

3 0

Response:

There is no correlation

Detailed explanation:

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Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a

zzz [12365]

Answer:

Step-by-step explanation:

It has been established that the count of drivers traveling between a specific origin and destination in a certain time frame follows a Poisson distribution with a mean μ = 20 (as indicated in the article "Dynamic Ride Sharing: Theory and Practice"†).

a) $P(X\leq 15) = 0.1565=0.157$

b) $P(X>26) =1-F(26)\\= 1-0.9221\\=0.0779=0.078$

c) $P(15\leq x\leq 26)\\=F(26)-F(14)\\=0.9221-0.1049\\=0.8172=0.817$

d) 2 standard deviations = 2(20) = 40

Thus, this means the range for 2 standard deviations is

20-40, 20+40

which equates to (0,60)

$P(0$

5 0

3 months ago

Which expressions are equivalent to RootIndex 3 StartRoot 128 EndRoot Superscript x? Select three correct answers.

AnnZ [12381]

Answer:

- 128 Superscript StartFraction 3 Over x EndFraction

- (4RootIndex 3 StartRoot 2 EndRoot)x

- (4 (2 Superscript one-third Baseline) ) Superscript x

Step-by-step explanation:

Considerando la ecuación dada $(\sqrt[3]{128} )^{x}\\$

De acuerdo con una de las leyes de índices,

$(\sqrt[a]{m} )^{b}\\= (\sqrt{m})^\frac{b}{a}$

Aplicando esta ley a la pregunta;

$(\sqrt[3]{128} )^{x}\\ = {128} ^\frac{x}{3}\\ \\= (\sqrt[3]{64*2})^{x} \\ = (4\sqrt[3]{2})^{x} \\= (4(2^{1/3} )^{x} )$

<pLos siguientes son ciertos de acuerdo con el cálculo presentado

128 Superscript StartFraction 3 Over x EndFraction

(4RootIndex 3 StartRoot 2 EndRoot)x

(4 (2 Superscript one-third Baseline) ) Superscript x

5 0

3 months ago

Read 2 more answers

Mike is 5 years old and Randy is 8 years old. Select all proportional relationships that have the same ratio to that of Mike and

lawyer [12517]

Response:

Explanatory steps:

I am unsure as well. However, I can state that the first one is

Sarah is 40 years old and her mother is 64 years old.

7 0

3 months ago

A pebble is tossed into the air from the top of a cliff. The height, in feet, of the pebble over time is modeled by the equation

PIT_PIT [12445]

Answer:

96

Step-by-step explanation:

The highest height achieved by the pebble represented by the quadratic function

$h(t)=-16t^2+32t+80$ can be determined by finding the vertex.

First, we identify the t-coordinate of this vertex. This maximum height corresponds to this value of t, thus, we need to evaluate the h(t)-coordinate.

By comparing $h$ and $at^2+bt+c$ , we observe that:

$a=-16$

$b=32$

$c=80$

Next, we compute to uncover the t-coordinate of the vertex:

$t=\frac{-b}{2a}$

$t=\frac{-32}{2(-16)}$

$t=\frac{-32}{-32}$

$t=1$

Now, substituting t into $-16t^2+32t+80$ with 1 will enable us to find the corresponding h(t)-coordinate:

$-16(1)^2+32(1)+80$

$-16(1)+32+80$

$-16+32+80$

$16+80$

$96$

5 0

2 months ago

Pepita and francisco each spend a equal amount of time on homework. The table shows the fraction of time they spend on each subj

Zina [12379]

Answer:

$\begin{array}{ccc}\text{Homework}&\text{Pepita}&\text{Francisco}\\ \\\text{Math}&\dfrac{1}{6}&\dfrac{1}{2}\\ \\\text{English}&\dfrac{2}{3}&\dfrac{1}{8}\\ \\\text{Science}&\dfrac{1}{6}&\dfrac{3}{8}\end{array}$

Step-by-step explanation:

Pepita and Francisco allocate equal amounts of time to their homework. The table indicates how they distribute their time across various subjects.

$\begin{array}{ccc}\text{Homework}&\text{Pepita}&\text{Francisco}\\ \\\text{Math}&&\dfrac{1}{2}\\ \\\text{English}&\dfrac{2}{3}&\\ \\\text{Science}&\dfrac{1}{6}&\dfrac{3}{8}\end{array}$

Pepita dedicated

$\dfrac{2}{3}+\dfrac{1}{6}=\dfrac{2\cdot 2+1\cdot 1}{6}=\dfrac{5}{6}$

to English and Science.

Thus, she utilized

$1-\dfrac{5}{6}=\dfrac{6}{6}-\dfrac{5}{6}=\dfrac{1}{6}$

for Math.

On the other hand, Francisco spent

$\dfrac{1}{2}+\dfrac{3}{8}=\dfrac{1\cdot 4+3\cdot 1}{8}=\dfrac{7}{8}$

on Math and Science, resulting in him spending

$1-\dfrac{7}{8}=\dfrac{8}{8}-\dfrac{7}{8}=\dfrac{1}{8}$

on English.

The table displays:

$\begin{array}{ccc}\text{Homework}&\text{Pepita}&\text{Francisco}\\ \\\text{Math}&\dfrac{1}{6}&\dfrac{1}{2}\\ \\\text{English}&\dfrac{2}{3}&\dfrac{1}{8}\\ \\\text{Science}&\dfrac{1}{6}&\dfrac{3}{8}\end{array}$

5 0

2 months ago