1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nata0808 [166]
3 years ago
6

What is 255.89 rounded to 1 significant figure?can anybody help me with the rest?

Mathematics
1 answer:
mixas84 [53]3 years ago
5 0
I can help you with this one because I can't see the paper too well. Sorry. This one would be 256.
You might be interested in
Multiply.<br> 2 1/7 x 3 2/3<br><br> Answer with a mixed number in the simplest form
viva [34]

Answer:

7 6/7

Step-by-step explanation:

2\frac{1}{7} × 3\frac{2}{3} =

\frac{15}{7} × \frac{11}{3} =

\frac{165}{21} =

7\frac{6}{7}

4 0
2 years ago
5. What is the graph of the equation?<br> -5x + y=-3
Mariulka [41]

Answer:

Plot y=5x-3

Step-by-step explanation:

6 0
3 years ago
How to I solve problem 1?
Verdich [7]
#1

The uniforms are numbered 0, 1, 2, ..., 99. That's 100 numbers. Half of them are odd and half of them are even. So the probability that any one of the uniforms is odd is 1/2 just like the probability that any one uniform is even is 1/2.

(a) The numbers on the uniforms are independent of one another. That is, the number of her cross-country uniform does not in any way determine the number on her basketball uniform and vice versa. This means that we can find the probability that each is odd and multiply these together using what is called the counting principle. The probability that all are odd is:
(1/2)(1/2)(1/2)=1/8

(b) This is done the same way we did part (a). Since the probability of any one uniform being odd is the same as it being even (1/2), the answer here is the same: (1/2)(1/2)(1/2)=1/8

(c) This problem differs from that in (a) and (b). There is only one way for all three uniforms to be odd numbers: (odd, odd, odd) or all even (even, even, even). However, there are multiple ways for the uniforms to be two odd and one even. If the uniforms are listed in order: cross-country, basketball, softball we can get exactly one even in any of three ways:
even, odd, odd
odd, even, odd
odd, odd, even
The probability for any one of these possibilities is (1/2)(1/2)(1/2)=1/8 but since there are three way the probability that we get even exactly once is equal to (3)(1/8) = 3/8
7 0
3 years ago
A consulting firm submitted a bid for a large research project. The firm's management initially felt they had a 50-50 chance of
Verizon [17]

Answer:

Probability = 0.50

Step-by-step explanation:

Given

Probability = 50\ 50\ chance  

Successful\ Bids = 80\%

Unsuccessful\ Bids = 35\%

Required

Determine the probability before the request of past experience

In this case, we only consider the 50 50 chance probability

Probability = \frac{50}{50 + 50}

Probability = \frac{50}{100}

Probability = 0.50

5 0
3 years ago
Can someone pls help me I’m stuck
scZoUnD [109]

Answer:

i am sorry i have no idea sorry

6 0
3 years ago
Read 2 more answers
Other questions:
  • How do I construct angle XYZ with a measure of 2 times the measure of angle ABC
    5·1 answer
  • What does the ratio 5/1 represent in terms of the example.​
    10·1 answer
  • 17. A celebrity makes $0.05 for each like they get on
    9·2 answers
  • If secant theta equals 25/24 find cotangent theta
    15·2 answers
  • Find the value of each variable ​
    13·1 answer
  • How much time has passed from the first clock to the second clock?
    12·1 answer
  • Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
    14·1 answer
  • What is the area of a rectangle with vertices (2, 3), (9, 3), (9, Negative 2), and (2, Negative 2)?
    15·2 answers
  • 1. Q1) Which number is missing from this prime factor<br>tree?<br>​
    15·1 answer
  • In a certain store, the regular price of a refrigerator is$600. How much money is saved by buying this refrigeratorat 20 percent
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!