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3 years ago

Find the length of the missing segment

Mathematics

2 answers:

Bess [88]3 years ago

7 0

Answer:

18 or 15

Step-by-step explanation:

Mark me brainleist please

garri49 [273]3 years ago

4 0

Answer:

24, I just had the same exact question on my test lol

Step-by-step explanation:

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PLEASE HELP 35 POINTS MATH

ICE Princess25 [194]

Answer:

Step-by-step explanation:

-4+16+19=31

5 0

1 year ago

You work as a quality control inspector of shirts for a clothing manufacturing company. Today you examine one batch and 15% fail

Otrada [13]

Answer:

23%

Step-by-step explanation:

To find how many failed inspection, add together all the percentages:

15 + 5 + 3

= 23

So, 23% of the products failed inspection

5 0

2 years ago

‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️I need help ASAP 30 points and brainliest to correct answer and most helpful. Please elaborate and explain y

Strike441 [17]

1. Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.

What is the theoretical probability that the family has two dogs or two cats?

25% chance

2. Describe how to use two coins to simulate which two pets the family has.

You could use the coins to simulate which pet the family has by flipping them and having head be dog and tails be cat (or vice-versa).

3. Flip both coins 50 times and record your data in a table like the one below.

Based on your data, what is the experimental probability that the family has two dogs or two cats?

Based on the results, I concluded that for Heads, Heads (which could be dogs or cats) there was a 24% chance and for Tails, Tails there was a 26% chance

4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?

1/8 chance (accidentally messed up there) or 12.5%

5. How could you change the simulation to generate data for three pets?

To flip 3 coins and add more spots on the chart.

I hope that this helps because it took a while to write out. If it does, please rate as Brainliest

5 0

2 years ago

Read 2 more answers

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05

Tresset [83]

Answer:

(a) Probability that there are no surface flaws in an auto's interior is 0.6065 .

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws is 0.00673 .

Step-by-step explanation:

We are given that the number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05 flaws per square foot of plastic panel.

Let X = Distribution of number of surface flaws in plastic panels

So, X ~ Poisson( $\lambda$ )

The mean of Poisson distribution is given by, E(X) = $\lambda$ = 0.05

which means, X ~ Poisson(0.05)

The probability distribution function of a Poisson random variable is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}; for x=0,1,2,3...$

Now, we know that $\lambda$ for per square foot of plastic panel is 0.05 and we are given that an automobile interior contains 10 square feet of plastic panel.

Therefore, for 10 square foot of plastic panel is = 10 * 0.05 = 0.5

(a) Probability that there are no surface flaws in an auto's interior =P(X=0)

P(X = 0) = $\frac{e^{-0.5}*0.5^{0}}{0!}$ = $e^{-0.5}$ = 0.6065

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws = $P(X = 0)^{10}$

So, $P(X = 0)^{10}$ = $0.6065^{10}$ = 0.00673

8 0

3 years ago

A nationwide census is conducted and it is found that the mean number of hours of television watched per year by Americans is 35

Ber [7]

Answer:

Probability that a group of 4 Americans watch more than 400 hours of television per year is 0.3264.

Step-by-step explanation:

We are given that a nationwide census is conducted and it is found that the mean number of hours of television watched per year by Americans is 350 with a standard deviation of 220.

A group of 4 Americans is selected.

Let $\bar X$ = sample mean number of hours of television watched per year

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 350

$\sigma$ = standard deviation = 220

n = sample of Americans = 4

Now, the probability that a group of 4 Americans watch more than 400 hours of television per year is given by = P( $\bar X$ > 400 hours)

P( $\bar X$ > 400) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{400-350}{\frac{220}{\sqrt{4} } }$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

Hence, the probability that a group of 4 Americans watch more than 400 hours of television per year is 0.3264.

6 0

3 years ago