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mars1129 [50]
3 years ago
7

5. The box plot summarizes the test scores for 100 students: u 55 60 65 70 75 80 85 90 95 100 Which term best describes the shap

e of the distribution? a. bell-shaped C. skewed b. uniform d. symmetric​

Mathematics
1 answer:
ASHA 777 [7]3 years ago
7 0
C would be the correct answer
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Water fills a tank at a rate of 150 litres during the first hour, 350 litres during the second, 550 litres during the third and
Alborosie
Putting this as an arithmetic sequence gives:

u_n = 150+200(n-1)

The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L

The sum of an arithmetic series can be written as:

S_n=n/2 [2a+(n-1)d] = 784 000
\\n/2[2(150)+(n-1)200] = 784 000
\\n[300+200(n-1)=1 568 000
\\300n+200n^2-200n = 1 568 000
\\200n^2+100n- 1 568 000 = 0
\\2n^2 +n- 15680 = 0

\\n= 88.2...,-88.7

n has to be positive, so we get

n = <u>88.2 hours (3 s.f.)</u>
3 0
4 years ago
Read 2 more answers
Guys plz help Answer this QUESTION!!!
ycow [4]

Answer:

I beliEve it is D

Step-by-step explanation:

3 0
3 years ago
Please can someone help quickly
rodikova [14]
((5-2)*180)/5=SPT
SPT=108
360-(108+90)=RPT
RPT=162
((n-2)*180)/n=162
162n=(n-2)*180
.9n=n-2
.1n=2
n=20
8 0
3 years ago
BRAINLIESTTT ASAP!! PLEASE HELP ME :)
Andrej [43]

Answer:

2017.

Step-by-step explanation:

The equation for the future sale of motorcycles

= 5.75(1.16)^t  where t is in years

For cars it is

3.5(1.25)^t

3.5(1.25)^t > 5.75(1.16)^t

(1.25)^t > (5.75/3.5) 1.16^t

t ln 1.25 > ln 1.64286 + t ln 1.16

t ln 1.25 -  t ln 1.16 > ln 1.64286

t (ln 1.25 - ln 1.16) > ln 1.64286

t > ln 1.64286 / (ln1.25 - ln 1.16)

t > 6.64 .

So the answer is in the year 2017.

4 0
3 years ago
Read 2 more answers
Sally Sue had spent all day preparing for the prom. All the glitz and the glamour of the evening fell apart as she stepped out o
Nuetrik [128]

We have the function

p(t)=550(1-e^{-0.039t})

Therefore we want to determine when we have

p(t_0)=550

It means that the term

e^{-0.039t}

Must go to zero, then let's forget the rest of the function for a sec and focus only on this term

e^{-0.039t}\rightarrow0

But for which value of t? When we have a decreasing exponential, it's interesting to input values that are multiples of the exponential coefficient, if we have 0.039 in the exponential, let's define that

\alpha=\frac{1}{0.039}

The inverse of the number, but why do that? look what happens when we do t = α

e^{-0.039t}\Rightarrow e^{-0.039\alpha}\Rightarrow e^{-1}=\frac{1}{e}

And when t = 2α

e^{-0.039t}\Rightarrow e^{-0.039\cdot2\alpha}\Rightarrow e^{-2}=\frac{1}{e^2}

We can write it in terms of e only.

And we can find for which value of α we have a small value that satisfies

e^{-0.039t}\approx0

Only using powers of e

Let's write some inverse powers of e:

\begin{gathered} \frac{1}{e}=0.368 \\  \\ \frac{1}{e^2}=0.135 \\  \\ \frac{1}{e^3}=0.05 \\  \\ \frac{1}{e^4}=0.02 \\  \\ \frac{1}{e^5}=0.006 \end{gathered}

See that at t = 5α we have a small value already, then if we input p(5α) we can get

\begin{gathered} p(5\alpha)=550(1-e^{-0.039\cdot5\alpha}) \\  \\ p(5\alpha)=550(1-0.006) \\  \\ p(5\alpha)=550(1-0.006) \\  \\ p(5\alpha)=550\cdot0.994 \\  \\ p(5\alpha)\approx547 \end{gathered}

That's already very close to 550, if we want a better approximation we can use t = 8α, which will result in 549.81, which is basically 550.

Therefore, we can use t = 5α and say that 3 people are not important for our case, and say that it's basically 550, or use t = 8α and get a very close value.

In both cases, the decimal answers would be

\begin{gathered} 5\alpha=\frac{5}{0.039}=128.2\text{ minutes (good approx)} \\  \\ 8\alpha=\frac{8}{0.039}=205.13\text{ minutes (even better approx)} \end{gathered}

7 0
1 year ago
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