17/4 minutes with 510 total beats How many beats per minute

The masses of the oranges on sale at a farm stand are normally distributed with a mean of 239 grams and a standard deviation of

AleksAgata [21]

In order to calculate the z-score of a normal distribution, you need to apply the formula:

z = (o - m) / s

where:
o = observed value
m = mean
s = standard deviation

In your case:

z = (259 - 239) / 25 = 0.8

The answer is, indeed, z = 0.8

4 0

3 years ago

What are the factors of 37, 51, and 45? i just dont get it

o-na [289]

45=1,3,5,9,15,45 37=only 1 and 37 because it is prime. 51= 1,3,17,51

3 0

3 years ago

Complete the proof of the exterior angle theorem. Given: ∠4 is an exterior angle of △ABC. Prove: m∠1+m∠2=m∠4 Answer: Statement R

Nataly_w [17]

Answer:

Step-by-step explanation:

Statement Reason

1. ∠4 is an exterior angle of ΔABC Given

2. ∠3 and ∠4 form a linear pair Linear pair property

3. ∠3 is supplementary to ∠4 Straight line property

4. m∠3+m∠4=180° ∠3 is supplementary to ∠4

5. m∠1+m∠2+m∠3=180° Triangle sum theorem

6.m∠1+m∠2+m∠3= m∠3+m∠4 Statement 4 and 5

7. m∠1+m∠2=m∠4 Subtraction property

5 0

3 years ago

Read 2 more answers

The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.

blagie [28]

Answer:

$N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}$

Where $t_{1/2}= 3.3 hr$ represent the half life and the intial amount would be $N_o = 1$

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

$0.05 = 1 (0.5)^{t/3.3}$

If we apply natural log on both sides we got:

$ln(0.05) = \frac{t}{3.3} ln (0.5)$

And solving for t we got:

$t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26$

So then would takes about 14.26 hours in order to have 95% of the lead to decay

Step-by-step explanation:

For this case we can define the variable of interest amount of Pb209 and for the half life would be given:

$N(t) =N_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}$

Where $t_{1/2}= 3.3 hr$ represent the half life and the intial amount would be $N_o = 1$

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

$0.05 = 1 (0.5)^{t/3.3}$

If we apply natural log on both sides we got:

$ln(0.05) = \frac{t}{3.3} ln (0.5)$

And solving for t we got:

$t= 3.3 *\frac{ln(0.05)}{ln(0.5)}= 14.26$

So then would takes about 14.26 hours in order to have 95% of the lead to decay

4 0

3 years ago

Pentagon RSTUV is circumscribed about a circle. What is the value of x if RS = 6, ST = 9, TU = 7, UV = 15, and VR = 14?

Ulleksa [173]

I think the answer is uv=15

8 0

4 years ago