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
docker41 [41]
2 years ago
6

Solve - 4x + 2 2 26. O A. X>6 O B. X<6 O C. X>6 O D. X<-6

Mathematics
1 answer:
vichka [17]2 years ago
7 0

Answer:

Step-by-step explanation:

This is quit complicated

You might be interested in
PLSSSSSS HELP MEE ASAP
schepotkina [342]

Answer:

What's the question?

Step-by-step explanation:

Your struggling with

7 0
2 years ago
A cake company makes cheesecakes with a circumference of approximately 75.36 cm. They need to design their boxes to offer single
Umnica [9.8K]

Answer:

12

Step-by-step explanation:

I'm not entirely sure abt this but here goes nothing

75.36 divided by 3.14 = 24

24 = diameter

radius = 12

7 0
2 years ago
Read 2 more answers
The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g
Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 \\ n \geq 30. Mathematically

\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean \\ \mu = 0 and a population standard deviation \\ \sigma = 1.

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

From the question, we know that

  • The population mean is \\ \mu = 88 WPM
  • The population standard deviation is \\ \sigma = 14 WPM

We also know the size of the sample for this case: \\ n = 137 sixth graders.

We need to estimate the probability that a sample mean being greater than \\ \overline{X} = 89.87 WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}

\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}

\\ Z = 1.5634 \approx 1.56

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that \\ P(\overline{X} < 89.87). Then

\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

5 0
3 years ago
Michaels buys three over a pound of cheese he puts the same amount of cheese and three sandwiches how many cheese how much chees
Tasya [4]

Question is not proper,Proper question is given below;

Nicholas buys 3/8 pound of cheese he put the same amount of cheese and three sandwiches how much cheese does Nicholas put on each sandwich

Answer:

Nichole can put \frac{1}{8} \ pounds of cheese on each of his sandwich.

Step-by-step explanation:

Total Amount of cheese he has = \frac{3}{8} \ pound

Number of cheese sandwiches = 3

We need to find Amount of cheese does Nicole put on each of his sandwich.

Solution:

Amount of cheese on each sandwich can be calculated by dividing Total Amount of cheese he has with Number of cheese sandwiches.

framing in equation form we get;

Amount of cheese on each sandwich = \frac{\frac{3}{8}}{3}} = \frac{1}{8} \ pounds

Hence Nichole can put \frac{1}{8} \ pounds of cheese on each of his sandwich.

7 0
3 years ago
Find the 7th term of the arithmetic sequence 4x + 5, 11x + 9,18x + 13, ...
Svet_ta [14]

Step-by-step explanation:

we see the common difference is

a2-a1 = 11x + 9 - 4x - 5 = 7x + 4

an = an-1 + (7x+4) = a1 + (n-1)×(7x+4)

for a7 we have then

a7 = 4x + 5 + 6×(7x +4) = 4x + 5 + 42x + 24 = 46x + 29

5 0
2 years ago
Other questions:
  • Does 24, 7, and 26 form a right triangle
    13·1 answer
  • If there was a circle and i wanted to find out the area.
    5·1 answer
  • Fill in the blank to make the following a true statement.
    12·1 answer
  • What are the missing angles?
    8·1 answer
  • Por favor ayúdeme!!​
    15·1 answer
  • Pls this is urgent<br><br>A. f(x)=4x-3<br>B.f(x)=2x+3<br>C.f(x)=3x+1<br>D.f(x)=3x-1​
    5·1 answer
  • 5v-15=20<br> solve for v<br> can anyone explain what im supposed to do
    15·2 answers
  • One of the legs of a right triangle measures 5 cm and its hypotenuse measures 9 cm.
    5·1 answer
  • A.CED.A Garnetta owned 2400 shares of Metropolitan Corporation at a price of $45.50. The stock split 4-for-3. How was Garnetta f
    11·1 answer
  • What is the maximum value for y in the following function?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!