__________ notation encodes a real number as a mantissa multiplied by a power (exponent) of 2.

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ 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)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, 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:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (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".

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

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0

2 years ago

HELP, PLZ FAST WILL GIVE BRAINLIEST ANSWER!!!!!!!!!!!!!!!!

kotykmax [81]

The answer will be 200 inches. 508 ÷ 2.54 = 200

4 0

2 years ago

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PLEASE HELP<br><br> Find the surface area of the triangular prism.

ASHA 777 [7]

Answer:

139

Step-by-step explanation:

First, find all the faces that you must find the areas of. In this case, you need to find the areas of two triangles and three rectangles.

In this triangular prism, the base is one triangle. If the area of the base is 7.75, we can assume that that holds true for both bases, and multiply 7.75 by 2 in order to find the area of both triangles.

Now find the area of each of the three rectangles by multiplying their individual heights and bases by each other. You should get 38, 47.5, and 38 as the areas of the rectangles.

Now add all the individual area's together. (They're bolded for clarity).

4 0

2 years ago

Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin

Airida [17]

Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

$N(t)=N _{0} e^{- \alpha t}$ ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) = No / 2

So, replace in the given equation:

$N_{t-half} = N_{0} /2 = N_{0} e^{- \alpha t}$

3) Solving for α (remember α is λ)

$\frac{1}{2} = e^{- \alpha t} 

2 = e^{ \alpha t} 

 \alpha t = ln(2)$

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ

4 0

2 years ago

Square roots question??

cluponka [151]

√5²+2*√5*√3+√3²

5+7.7+3

15.7

3 0

2 years ago

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