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2 years ago

11

.A teacher has 70 markers inter drawer. She

1 answer:

AlekseyPX2 years ago

4 0

Answer:

20%

Step-by-step explanation:

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joy added 26 new contacts to her phone list. She now has a total of 100 contacts. Let c represent how many contacts joy had on h

Nesterboy [21]

Answer: 26 + c = 100

Step-by-step explanation: To find what c equals, we must flip the solution around and subtract it. When we take 100 - 26, we get 74. Therefore, c = 74. I hope this helped! :)

7 0

2 years ago

Determine whether (7,9) is a solution of 7x+2y= −7.

topjm [15]

We can check by subbing in the values.

7x + 2y = -7
(7,9).....x = 7 and y = 9
now we sub
7(7) + 2(9) = -7
49 + 18 = -7
67 = -7 (incorrect)....therefore, (7,9) is NOT a solution for this equation

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3 years ago

Solve for this equation, will mark brainliest for correct answers

cluponka [151]

Answer:I think its D

Step-by-step explanation:

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2 years ago

Read 2 more answers

Write a rule that represents the function. <br> (1,1/2),(2,1/4),(3.1/8),(4,1/16),(5,1/32)

ratelena [41]

Answer:

Step-by-step explanation:

. cubesi, 20,64. AV 1,6,11,16... Add 5. 1, 2, 4, 8. . Multiply 2. $2,9,28,65... cubes add 1. 31,49,16 ...

5 0

2 years ago

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the

Sergio039 [100]

Answer:

The estimate is $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is $\r p__{hat}} = 0.37$

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

$\alpha = 100 - 90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=> $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=> $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

$P__{hat}} \pm E$

substituting values

$0.37 \pm 0.0348$

5 0

2 years ago