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
kvv77 [185]
2 years ago
15

I need some help with zee maths.

Mathematics
1 answer:
seraphim [82]2 years ago
8 0
You have to collect like terms that means group the numbers that have the same variables

You might be interested in
Help me find the area :)
Andreas93 [3]

Answer:

20.625 ft²

(hope this is right :)

4 0
2 years ago
A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s
Umnica [9.8K]

Answer:

88.51 is the minimum score needed to receive a grade of A.          

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.0793.

P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793  

= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793  

=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207  

Calculation the value from standard normal z table, we have,  

P(z \leq 1.410) = 0.9207

\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51  

Hence, 88.51 is the minimum score needed to receive a grade of A.

3 0
2 years ago
Phonics is an instructional method in which children are taught to connect sounds with letters or groups of letters. A sample of
Fantom [35]

Answer:

A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645\frac{23.44}{\sqrt{141}}

M = 3.25

The lower end of the interval is the sample mean subtracted by M. So it is 34.09 - 3.25 = 30.84.

The upper end of the interval is the sample mean added to M. So it is 34.09 + 3.25 = 37.34.

A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).

6 0
3 years ago
What values of x make the equation x2 + 9x – 22 = 0 true? Check all that apply.
Inga [223]
-11 and 2 make the equation true. 
3 0
3 years ago
Read 2 more answers
Can someone answer this? Thanks a lot
Komok [63]

Answer:

Step-by-step explanation:

Use math-way go quick and easy answers without step by step explanations, and use math-soup-calculator for step by step explanation

3 0
3 years ago
Read 2 more answers
Other questions:
  • Help?
    8·2 answers
  • What is five and five sixths divided by five twelfths?
    6·2 answers
  • Please help me and hurry
    10·2 answers
  • Read the following step of solving a System of Equations problem and explain if it is correct or incorrect.
    15·1 answer
  • -2/5 + 1/2 please help guys I really need it
    13·1 answer
  • Which expressions are equivalent to the on below select all that apply 9x
    15·2 answers
  • Which one of these statements is correct????
    12·1 answer
  • Help me please this is due at 11:30
    7·2 answers
  • As Sales Manager for ISeeYou Productions, you are planning to review the prices you charge clients for television advertisement
    7·1 answer
  • How do I find the period, amplitude, midline, and vertical shift. I'm very confused.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!