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Thepotemich [5.8K]
3 years ago
9

10>x/3+6 help me please

Mathematics
2 answers:
aleksley [76]3 years ago
4 0
Minus 6 both sides
4>x/3
times3 both sides
12>x
Dennis_Churaev [7]3 years ago
3 0
Subtract 6 from both sides:
10−6>x/3

Simplify <span>10-6<span>10−6</span></span><span> to </span><span>4<span>:
</span></span>4><span><span>​3</span>​<span>​<span>x

</span></span></span>Multiply both sides by 3:
<span>4×3>x
</span><span>
Simplify </span><span>4\times <span>4×3</span></span><span> to </span><span>12:
</span>12><span>x
</span>
switch the inequality sign:
<

Answer:
x<12

hope i helped

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original price of a calendar: $14.50 discount: 30% find the new price for the item after the discount
tigry1 [53]
Here, Original price = $14.50
Discount = 30%

Now, Amount of discount = 14.50*0.30 = $4.35

So, price after discount would be: $14.50 - $4.35 = $10.15

Your answer is $10.15

Hope this helps!
5 0
3 years ago
Read 2 more answers
Juanita and Nina are bowling together. The probability of Juanita getting a strike next game is 24%. The probability of Nina get
ASHA 777 [7]

Answer:

(Choice A)

Juanita gets a strike next game.'

Step-by-step explanation:

Your question is obviously incomplete.

Complete question is:

Juanita and Nina are bowling together. The probability of Juanita getting a strike next game is 24%. The probability of Nina getting a strike next game is 0.17. Which of these events is more likely?

(Choice A)

Juanita gets a strike next game.'

(Choice B)

Nina gets a strike next game.

(Choice C)

Neither. Both events are equally likely.

Answer:

Probability of Juanita : P(J)= 24% => 0.24

probability of Nina getting a strike next game: P(N) = 0.17

As you can see 0.24 > 0.17 ----> P(J)>P(N)

Thus it can be concluded that Juanita gets a strike next game is more likely.

so, choice A

4 0
3 years ago
Read 2 more answers
At a​ school, 118 students play at least one sport. This is 40​% of the students at the school. How many students are at the​ sc
REY [17]
There are 295 students.
4 0
3 years ago
What exact area of a circle with a radius of 6 feet?
Volgvan

Answer:


Step-by-step explanation:

3.14(pi) X 6 ^2 =

3.14*6 = 18.84

3.14x6^2 =36

your final answer is : B. : 36

7 0
3 years ago
During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for
AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is :  \hat p = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

      P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion = 0.345

           n = sample of participants = 3532

           p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population​ proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                         significance are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u>= [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

    = [ 0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } , 0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } ]

    = [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

6 0
3 years ago
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