What is the order of rotational symmetry for a mombus? 0 1

Car a and b are travelling at 30 and 40 mph. if car b starts out 5 miles behind car 6, how many hours will it take for car b to

Leviafan [203]

As both are moving in the same direction ,both will travel the same distance till catch up time --d

4 0

3 years ago

7)Twice a number when decreased by 7 gives 45. Find the number.

kirza4 [7]

Isn’t it 26... cuz 26x2 is 52 then subtract 7 which is 45

4 0

2 years ago

the atlanta braves are offering bobblehead to every 10th person who enters your fundraiser. the nashville predators are offering

Paha777 [63]

In this situation, attendee number thirty will be the first one o receive both the bobblehead and the ticket.

<h3>Which attendees will receive the bobblehead and the ticket?</h3>

Bobblehead:

Attendee number 10
Attendee number 20
Attendee number 30

Ticket:

Attendee number 15
Attendee number 30

<h3>What can be concluded?</h3>

It can be concluded the attendee number 30 will be the first one to receive both the ticket and the bobblehead.

Learn more about ticket in: brainly.com/question/14001767

#SPJ1

6 0

1 year ago

Each item produced by a certain manufacturer is independently of acceptable quality with probability 0.95. Approximate the proba

Diano4ka-milaya [45]

Answer:

The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

Step-by-step explanation:

Let X = number of items with unacceptable quality.

The probability of an item being unacceptable is, P (X) = p = 0.05.

The sample of items selected is of size, n = 150.

The random variable X follows a Binomial distribution with parameters n = 150 and p = 0.05.

According to the Central limit theorem, if a sample of large size (n > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.

The mean of this sampling distribution is: $\mu_{\hat p}= p=0.05$

The standard deviation of this sampling distribution is: $\sigma_{\hat p}=\sqrt{\frac{ p(1-p)}{n}}=\sqrt{\frac{0.05(1-.0.05)}{150} }=0.0178$

If 10 of the 150 items produced are unacceptable then the probability of this event is:

$\hat p=\frac{10}{150}=0.067$

Compute the value of $P(\hat p\leq 0.067)$ as follows:

$P(\hat p\leq 0.067)=P(\frac{\hat p-\mu_{p}}{\sigma_{p}} \leq\frac{0.067-0.05}{0.0178})=P(Z\leq 0.96)=0.8315$

*Use a z-table for the probability.

Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

5 0

3 years ago

What fraction of 3/4 it will take to make 2/3

Marizza181 [45]

3/4x = 2/3

4/3(3/4)x = (2/3)(4/3)

x = 8/9

4 0

2 years ago