Pls somone help me plzzzz

HELP what is 1/3 x 575?(This is hard for me)

Nonamiya [84]

Answer:

1/3 × 575 = 575/3

8 0

3 years ago

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per nig

Kitty [74]

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let $\mu$ denotes the average hours of sleep per night.

As per given , we have

$H_0:\mu=7\\H_a:\mu$

, since $H_a$ is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : $\overline{x}=7.24$

Sample standard deviation : s= 1.93

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

i.e. $t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58$

For significance level $\alpha=0.05$ and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= $t_{\alpha, df}=t_{0.05,21}=- 1.7207$

Decision : Since the test statistic value (0.58) > critical value 1.7207, it means we are failed to reject the null hypothesis .

[Note : When $|t_{cal}|>|t_{cri}|$ , then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

6 0

3 years ago

There are five nickels and five dimes in your pocket. With replacing, what is the probability that you will pull a nickel the fi

N76 [4]

Answer:

1/8

Step-by-step explanation:

Probability of pulling a nickel the first time: 5/10

Probability of pulling a nickel the second time: 5/10 (remember that you are replacing)

Probability of pulling a dime the third time: 5/10

Total probability: 5/10 * 5/10 * 5/10 OR 1/2 * 1/2 * 1/2 = 1/8

im pretty sure this is the answer! i hope this helps! <3

4 0

2 years ago

The function f(h)=m(1/2)^h gives the mass, m, of a radioactive substance remaining after h half-lives. Cobalt-60 has a half-life

goldfiish [28.3K]

Answer:

Third choice:

$f(x)=50(0.877)^{10};13.5mg$

Explanation:

1. Data:

Function:

$f(h)=m(1/2)^h$

Cobalt-60's half-life: 5.3 years

Initial mass of cobalt-60: 50 mg

2. Unknown:

equation for the mass of cobalt-60 remaining after 10 years = ?

mass remaining =?

3. Solution

Half-life is the time it takes a sample to decay to half of its initial amount. It is considered constant. Hence, when one half-life passes, the sample has decayed to 50% of the original amount; when two half-lives pass, the sample has decayed to (1/2)×(1/2) = 1/4 = 25%; when three half-lives have elapsed, the sample has decayed to (1/2)³ = 1/8 = 12.5% of its original amount, and so on.

Then, the amound of a sample remaining is calculated as the original amount times (1/2) raised to the number of half-lives elapsed, which is what the given function, $f(h)=m(1/2)^h$ models.

You just must substitute the data into the function to get the answer to the question:

$f(x)=50(0.5)^{(10/5.3)}$

Where, 50 is the original mass of 50g, 0.5 is equal to 1/2, and 10/5.3 gives the number of half-lives (the number of times that 5.3 years is contained in 10 years).

Simplifying:

$f(x)=50(0.5)^{(10/5.3)}=50((0.5)^{(1/5.3})^{10}\approx50(0.877)^{10}$