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3 years ago

9

A box holds 24 containers of yogurt. Each box has 4 flavors with 6 containers of each flavor. If you choose three yogurts withou

t looking and don't replace them, what is the probability of choosing three of the same flavor

1 answer:

zavuch27 [327]3 years ago

3 0

Answer:

3/6 or 1/2 (50%)

Step-by-step explanation:

There is 6 containers of each flavor, correct?

If you pick 3 of the same flavor, 3/6, it would look like that fraction.

If you simplify it, It'd be 1/2.

If you want it as a percent, It'd be 50%.

As a decimal, It's .50.

You might be interested in

Given △XPS≅△DNF, find the values of x and y.

White raven [17]

Answer:

x = 3

y = 15

Step-by-step explanation:

If △XPS ≅△DNF, their corresponding sides would be congruent. This implies that:

XP ≅ DN

PS ≅ NF

XS ≅ DF

Given that:

XP = 4y - 3

DN = 57

NF = 51

XS = 17x + 3

DF = 54

Therefore:

XP = DN

4y - 3 = 57 (Substitution)

Add 3 to both sides

4y = 57 + 3

4y = 60

Divide both sides by 4

y = 60/4

y = 15

Also,

XS = DF

17x + 3 = 54 (substitution)

Subtract 3 from each side

17x = 54 - 3

17x = 51

Divide both sides by 17

x = 51/17

x = 3

4 0

3 years ago

To make the perfect orange color hector mixes 2/3 quarts of red paint with 1 4th Court of yellow paint. How much red paint is us

jonny [76]

6/12 or the same thing but reduced 1/2

WELCOME!

5 0

3 years ago

Read 2 more answers

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

3 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

5 0

2 years ago

Solve for x . Leave your answer in the simplest form

Alisiya [41]

Hello!

Here we are given a composite shape, meaning that we can divide this shape up into ways that we are given formulas to solve for legs and such.

I can see here that this can be divided into a triangle and a rectangle using a horizontal line.

The leg lengths of this triangle would be 9, because it is congruent to the bottom side of the rectangle, and 7, because it is the left side minus the right side.

This would also make a right triangle, meaning we could solve for the hypotenuse utilizing the Pythagorean Theorem.

$a^2+b^2=c^2$

$6^2+7^2=c^2$

$36+49=c^2$

$c^2=85$

$c=\sqrt{85}$

Which is in simplest radical form.

Hope this helps!

3 0

2 years ago