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3 years ago

Pedro has determined that the probability his shot will score in a lacrosse game is 0.30.

Mathematics

2 answers:

Igoryamba3 years ago

8 0

For two shots Pedro has four outcomes:

score | score

score | not score

not score | score

not score | not score

The probability Pedro's shot will score in a lacrosse game is 0.30 and the probability his shot will not score in a lacrosse game is 1-0.30=0.70. So you can count probabilities for all cases:

1. 0.3·0.3=0.09;

2. 0.3·0.7=0.21;

3. 0.7·0.3=0.21;

4. 0.7·0.7=0.49.

In total 0.09+0.21+0.21+0.49=1. The first outcomes is that what you need.

Answer: 0.09.

bija089 [108]3 years ago

4 0

If the probability of making a shot is 3/10, the probability of making two shots in a row is:

(3/10)(3/10)=9/100 (9%)

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rjkz [21]

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Step-by-step explanation:

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3 years ago

In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t an

Alenkasestr [34]

Answer: 693 years

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

$A(t) =A_0e^{kt}$

k = rate constant

t = time taken for decomposition = 1

$A_0$ = Initial amount of the reactant

$A_t$ = amount of the reactant left = $A_0-\frac{0.1}{100}\times A_0=0.999A_0$

$0.999A_0=A_0e^{k\times 1}$

$0.999=e^k$

$k=-0.001year^{-1}$

for half life : $t=t_\frac{1}{2}$

$A_t=\frac{1}{2}A_o$

Putting in the values , we get

$\frac{1}{2}A_0=A_0e^{-0.001\times t_\frac{1}{2}$

$t_\frac{1}{2}=693 years$

Thus half life of this isotope, to the nearest year is 693.

8 0

3 years ago

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Burka [1]

Answer:

Step-by-step explanation:

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3 years ago

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4 sisters share 3 muffins equally how much does each get

Allushta [10]

I think that the 4 sisters get 3 pieces each of the muffins because if you cut it in four pieces it gives one piece to each one of the four sisters and you keep doing it with the other two left muffins.

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3 years ago

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A researcher records the repair cost for 8 randomly selected washers. A sample mean of $60.46 and standard deviation of $18.36 a

Anuta_ua [19.1K]

Answer:

The critical value is T = 1.895.

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 6

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.895, which is the critical value.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.895\frac{18.36}{\sqrt{8}} = 12.301$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 60.46 - 12.301 = $48.159

The upper end of the interval is the sample mean added to M. So it is 60.46 + 12.301 = $72.761

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761

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3 years ago