A. What is the minimum value taken by the expression (x

Tell whether the expressions in each pair are equivalent:p(4-4) and 0

kifflom [539]

P(4-4)=p(0)=0
0=0
true, they are equivilent

7 0

3 years ago

Strontium-90 has a half-life of 28.8 years. If you start with a 300-gram sample of strontium-90,

Kazeer [188]

Answer: 2.66 × 10⁻¹³

Step-by-step explanation:

First, use the decay formula $A=A_oe^{kt}$ where

A is the final amount (amount left)
A₀ is the initial amount (amount you started with)
k is the rate of decay (you need to solve for this)
t is the time

Given:

A = 1/2(300) = 150
A₀ = 300
k = unknown
t = 28.8

$150=300e^{28.8k}\\\\0.5=e^{28.8k}\\\\ln(0.5)=ln(e^{28.8k})\\\\ln(0.5)=28.8k\\\\\\\dfrac{ln(0.5)}{28.8}=k\\\\\\\large\boxed{-0.0240676=k}\\$

Next, input the k-value and the new t-value to solve for A.

A = unknown
A₀ = 300
k = -0.0240676
t = 1440

$A=300e^{1440(-.0240676)}\\\\\large\boxed{A=2.66\times 10^{-13}}\\$

5 0

4 years ago

The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 i

fenix001 [56]

Answer:

Null hypothesis: $\mu \geq 250$

Alternative hypothesis: $\mu < 250$

$p_v =P(t_{15}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=510$ represent the mean for the sample

$s=50$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =530$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$

4) Calculate the critical value

We need to begin calculating the degrees of freedom

$df=n-1=16-1=15$

The critical value for this case would be :

$P(t_{15}$

The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75

5) Calculate the P-value

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}$

6) Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

8 0

3 years ago

What is the absolute value of -8? Use the number line to help answer the question

Ad libitum [116K]

The absolute value of -8 is 8
here’s a trick the absolute value of a positive is the same number of its negative it’s the same number but positive
hope it helps

4 0

3 years ago

...............2+2-2+2+2-4+2

kirill115 [55]

Hi your answer is 4..

Please follow me..

8 0

3 years ago

Read 2 more answers

A. What is the minimum value taken by the expression (x – 4)2 + 6?