How to solve this math problem

4 employees drive to work in the same car. the workers claim they were late to work because of a flat tire. their managers ask t

Brums [2.3K]

Answer:

The required probability is 0.015625.

Step-by-step explanation:

Consider the provided information.

There number of tires are 4.

The number of workers are 4.

The probability that individual select front drivers side tire = $\frac{1}{4}$

The probability that they all select front drivers side,tire is: $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}=\frac{1}{256}$

Similarly,

The probability that they all select front passenger side, tire is: $\frac{1}{256}$

The probability that they all select rear driver side,tire is: $\frac{1}{256}$

The probability that they all select rear passenger side, tire is: $\frac{1}{256}$

Hence, the probability that all four workers select the same tire = $\frac{1}{256}+ \frac{1}{256}+\frac{1}{256}+\frac{1}{256}$

the probability that all four workers select the same tire = $\frac{4}{256}=\frac{1}{64}=0.015625$

Hence, the required probability is 0.015625.

5 0

3 years ago

42 is 8% of what number

Setler [38]

8%=0.08
42÷0.08=525
So your answer is 525

7 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

p values for chi-square tests provide a quantitative measure of the probability that the results of another replicate experiment

nlexa [21]

Answer:

0.05

Step-by-step explanation:

P value that is less than 0.05, that is P < 0.05 means that the null hypothesis is false and should be rejected. If the P value is greater than 0.05, that is P > 0.05, it means the null hypothesis is true. The P value is also called the calculated probability value. P value subtracted from 1 is the probability that the alternative hypothesis is true.

8 0

3 years ago

2 Days Left Please Help.) The table shows the test scores and the sleep averages of several students. A) Write the least squares

ehidna [41]

Answer:

It looks like you are cheating on your Algebra exam!

Step-by-step explanation:

8 0

3 years ago