How many solutions can a consistent system of linear equations have?

pochemuha

it's the only one that hasn't been with a positive 2

4 0

2 years ago

The sugar content of the syrup is canned peaches is normally distributed. Assumethe can is designed to have standard deviation 5

andreyandreev [35.5K]

Answer: 0.50477

Step-by-step explanation:

Given : The sugar content of the syrup is canned peaches is normally distributed.

We assume the can is designed to have standard deviation $\sigma=5$ milligrams.

The sampling distribution of the sample variance is chi-square distribution.

Also,The data yields a sample standard deviation of $s=4.8$ milligrams.

Sample size : n= 10

Test statistic for chi-square : $\chi^2=\dfrac{s^2(n-1)}{\sigma^2}$

i.e. $\chi^2=\dfrac{(4.8)^2(10-1)}{(5)^2}=8.2944$

Now, P-value = $P(\chi^2>8.2944)=0.50477$ [By using the chi-square distribution table for p-values.]

Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.50477

8 0

3 years ago

Which ten thousand does 117,290 round to

Leni [432]

Answer:

120,000

Step-by-step explanation:

because seven in greater than 5 so it would round up

3 0

3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago

What is the value of f=(x)=16x when x= 1/2

zalisa [80]

For the second question, your first option is correct, you would rewrite 8 in power-base form 8=2^3 as the first step to solving the equation, power base form is exponential form.

4 0

3 years ago