All categories

3 years ago

Which is equivalent to P(z ≥ 1.4)?

Mathematics

2 answers:

ohaa [14]3 years ago

4 0

Answer:

1st: B

2nd: 8%

3rd: 26%

Step-by-step explanation:

doing the assignment ;)

Bad White [126]3 years ago

3 0

Answer:

1-P(z less than or equal to 1.4)

Step-by-step explanation:

To know what sentence is equivalent to the expression P(z ≥ 1.4), we just need to know the translation of the symbol '≥'

This symbol means that the expression in the left side is greater than or equal to the expression in the right side, so we can translate this symbol as 'greater than or equal to'.

As P means a probability, we can also shift the symbol from 'greater than or equal to' to 'less than or equal to' if we make 1 minus the probability, so the expression P(z ≥ 1.4) is equivalent to 1 - P(z ≤ 1.4), and this second expression can be translated as 1-P(z less than or equal to 1.4)

You might be interested in

THE HCF OF TWO NUMBERS 2952 AND 2256 IS 24 FIND THEIR LCM<br>

MrRissso [65]

Step-by-step explanation:

by which method defination method, prime factorization or division method

5 0

2 years ago

The population of a certain species of oak tree in a national forest declines each year by one-fourth. So, after each year, only

dolphi86 [110]

Answer: 25%

Step-by-step explanation: The question tells us that the population decreases by 1/4 every single year. That means, since 1/4 equals .25, the population decreases by 25% every year :)

7 0

3 years ago

2. What are the solutions to the quadratic equation

Svetlanka [38]

Answer:

The answer is x=5 and x=−5

x² - 25 = 0

(x-5)(x+5)=0

x²-5x+5x-25=0

x-5=0 x+5=0

x=5 or x=-5

8 0

3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is: