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mezya [45]
3 years ago
15

Plz show your work. Thx!! 9m=6

Mathematics
1 answer:
grandymaker [24]3 years ago
5 0
So first you would divide the 9 on both sides then 6 divided by 9 would be .6 repeating so m=.6 repeating.

You might be interested in
32+2/3(21y-6) in factored form
artcher [175]

Answer:

14(y + 2).

Step-by-step explanation:

32 +2/3(21y - 6)

Distributing the 2/3 over the parentheses:

= 32 + 2/3*21y  + 2/3 * -6

= 32 + 14y - 4

= 14y + 28

= 14(y + 2).

6 0
3 years ago
Urgent math help needed!! Multiple choice question. Will mark brainliest!!!
muminat

Total number of 10 grader students = 260 students.

Total number of students who rides the bus and they are of 10th grader = 150 students.

Probability (a student rides the bus given that they are a 10th grader) =\frac{Total \ number \ of \ students \ rides \ the \ bus \ and \ 10th \ grader}{Total \ number \ of \ 10 \ grader \ students}

Therefore,

P(E) = \frac{150}{260}

Dividing top and bottom by 10, we get

P(E) = 15/26.

In decimals 0.577.

Therrefore, correct option is C) 150/260 = 15/26 = .577.

8 0
3 years ago
How many minutes does the white vacuum need to clean 86.25 square feet?
Dovator [93]

Answer:

1.5 hours

As a general rule, 1,000 square feet of the house should take 1.5 hours to clean, according to Cleaning 4 Profit. So a 3,000 square feet home should take 3 hours to clean, and so on.

6 0
2 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

8 0
3 years ago
Please tell me if its A,B,C,D, or E
kow [346]

Answer:

A

Step-by-step explanation:

I did the math! and got 15

Hope this helps!

6 0
2 years ago
Read 2 more answers
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