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1 year ago

Select the locations on the number line to plot the points −2 , −11 , and −17 .

1 answer:

8 0

Answer: With this information the answer would be GO down twice into the negatives and make you plot point, next go down eleven times and make your plot point, and last go down seventeen times and make your plot point.

Step-by-step explanation: What you have to do is make the number line with a zero at in the middle on the left of the zero you go down into the the negative numbers and on the right you go up into the positive numbers

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A small ferryboat is 4.00 m wide and 6.00 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.00 cm into t

Svet_ta [14]

Answer:

$F_w=9408\ N$

Weight of the truck=9408 N

Step-by-step explanation:

Boat is experiencing the buoyant force as it is in the water and is sinking

According to the force balance in y direction. As both is floating, two forces balance each other:

$F_b-F_w=0$

where:

$F_b$ is the buoyant force

$F_w$ is the weight=mg

$F_b=F_w$ Eq (1)

Buoyant force is equal to the mass of water displaced * gravitational acceleration.

$F_b=m_{water\ displaced}*g\\F_b=\rho Vg\\$

Taking density of water to be 1000 Kg/m^3

$F_b=1000*(6*4*4*10^{-2})*9.8\\F_b=9408 N$

From Eq(1):

$F_w=9408\ N$

Weight of the truck=9408 N

6 0

2 years ago

Automobile mechanics conduct diagnostic tests on 150 new cars of a particular make and model to determine the extent to which th

Aleks04 [339]

Answer:

99% Confidence interval: (0.185,0.375)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 150

Number of cars that have faulty catalytic converters, x = 42

$\hat{p} = \dfrac{x}{n} = \dfrac{42}{150} = 0.28$

99% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Putting the values, we get:

$0.28\pm 2.58(\sqrt{\frac{0.28(1-0.28)}{150}}) = 0.28\pm 0.095\\\\=(0.185,0.375)$

The 99% confidence interval for the true proportion of new cars with faulty catalytic converters is (0.185,0.375)

8 0

1 year ago

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in

ASHA 777 [7]

Answer:

a) There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c) The expected number of Chrome users is 4.074.

d) The variance for the number of Chrome users is 3.2441.

The standard deviation for the number of Chrome users is 1.8011.

Step-by-step explanation:

For each Internet browser user, there are only two possible outcomes. Either they use Chrome, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

Google Chrome has a 20.37% share of the browser market. This means that $p = 0.2037$

20 Internet users are sampled, so $n = 20$ .

a.Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is P(X = 8).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243$

There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b.Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

Either there are less than 3 Chrome users, or there are three or more. The sum of the probabilities of these events is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105$

$P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538$

$P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.1950$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1950 = 0.8050$

There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c.For the sample of 20 Internet browser users, compute the expected number of Chrome users

We have that, for a binomial experiment:

$E(X) = np$

$E(X) = 20*0.2037 = 4.074$

The expected number of Chrome users is 4.074.

d.For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

We have that, for a binomial experiment, the variance is

$Var(X) = np(1-p)$

$Var(X) = 20*0.2037*(0.7963) = 3.2441$

The variance for the number of Chrome users is 3.2441.

The standard deviation is the square root of the variance. So

$\sqrt{Var(X)} = \sqrt{3.2441} = 1.8011$

The standard deviation for the number of Chrome users is 1.8011.

6 0

2 years ago

56.66 divided by 3.4

eimsori [14]

16 .66

is the answer yess girl

5 0

1 year ago

The distributive property of 12(309) and the distributive property of 8(100+10+8)

ludmilkaskok [199]

1.3708
2. 800+80+64=944

3 0

1 year ago