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3 years ago

Write a standard form of a quadratic equation that has roots - 5/2 and 3.

Mathematics

1 answer:

lora16 [44]3 years ago

6 0

Answer:

2x^2 - x - 15 = 0.

Step-by-step explanation:

In factor form this is

(x + 5/2)(x - 3) = 0

x^2 + 5/2x - 3x - 15/2 = 0

x^2 - 1/2x - 15/2 = 0

Multiply through by 2:

2x^2 - x - 15 = 0.

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(5t^3)^-4<br><br> 1. 625/t^12<br> 2. 20/t^7<br> 3. 1/625t^12<br> 4. 1/20t^7

rusak2 [61]

Answer:

$\large\boxed{3.\ \dfrac{1}{625t^{12}}}$

Step-by-step explanation:

$\text{Use}\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\---------------\\\\(5t^3)^{-4}=\dfrac{1}{(5t^3)^4}=\dfrac{1}{5^4(t^3)^4}=\dfrac{1}{625t^{(3)(4)}}=\dfrac{1}{625t^{12}}$

4 0

3 years ago

The authors of a paper presented a correlation analysis to investigate the relationship between maximal lactate level x and musc

ArbitrLikvidat [17]

Answer:

a) Sample correlation coefficient, r = 0.7411

bi) test statistic, t = 4.102

bii) P-value = 0.000736

Step-by-step explanation:

a) The formula for the sample correlation coefficient is given by the formula:

$r = \frac{S_{xy} }{\sqrt{S_{xx} S_{yy} }} }$

$S_{xx} = 2,648,130.357\\S_{yy} = 36.7376,\\S_{xy} = 7408.225$

$r = \frac{7408.225}{\sqrt{2648130.357*36.7376} }$

r = 0.7511

i) formula for the test statistic is given by the formula:

$t = \frac{r\sqrt{n-1} }{\sqrt{1 - r^{2} } }$

sample size, n = 4

$t = \frac{0.7511\sqrt{14-1} }{\sqrt{1 - 0.7511^{2} } }$

t = 4.102

ii) Degree of freedom, df = n -2

df = 14 -2

df = 12

The P-value is calculate from the degree of freedom and the test statistic using excel

P-value =(=TDIST(t,df,tail))

P-value = (=TDIST(4.1,12,1)

P-value = 0.000736

4 0

3 years ago

How do you know which decmial is greater?

SOVA2 [1]

Answer:

Step-by-step explanation:

You have two decimals that both have the same number of digits in them. It is easy to compare decimals that have the same number of digits. Look at the numbers without the decimal point and determine which number is greater. 67 is greater.

6 0

3 years ago

A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa

UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so $\mu = \frac{14}{60} = 0.2333$