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inessss [21]
4 years ago
12

Please help me please

Mathematics
1 answer:
Volgvan4 years ago
3 0
The answer is 1.29k = t
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The price of the box of 10 markers is $7. The price of the box of 24 markers is $14. All prices without tax, and the price of th
stepladder [879]

Answer:

see below

Step-by-step explanation:

10 markers  =7 dollars

Divide each side by 10

10/10 = 7 /10

1 marker = 7/10

y =7/10 x  where y is the cost and x is the number of markers

24 markers  =14 dollars

Divide each side by 24

24/24 = 14 /24

1 marker = 7/12

y = 7/12 x  where y is the cost and x is the number of markers

6 0
3 years ago
A lot of 9 components contains 3 that are defective. Two components are drawn at random and tested. Let be the event that the fi
kirza4 [7]

Answer:

Probabilty of first component drawn defective= 0.1161

Probability of second component drawn defective= 0.1560

Step-by-step explanation:

Total component = 9

Defective component= 3

Non-defective component= 6

Probability of defective= 3/9

= 0.3333

Probabilty of non defective= 1-0.3333

= 0.6666

Probabilty of first component drawn is defective

= 9C1(0.333)¹(0.6666)^8

= 9(0.3333)(0.0387)

= 0.1161

Probabilty of second component is defective

= 8C1(0.3333)¹(0.6666)^7

= 8(0.3333)(0.0585)

= 0.1560

7 0
3 years ago
What are the values of x for which the denominator is equal to zero for y = x+3/x²+4x
Arisa [49]

Answer:

hello

x² + 4 x = 0

x ( x + 2  ) = 0

x = 0  or   - 2

7 0
3 years ago
In a random sample of 80 teenagers, the average number of texts handled in one day is 50. The 96% confidence interval for the me
Naily [24]

Answer:

Standard deviation of the sample = 17.421

Step-by-step explanation:

We are given that in a random sample of 80 teenagers, the average number of texts handled in one day is 50.

Also, the 96% confidence interval for the mean number of texts handled by teens daily is given as 46 to 54.

So, sample mean, xbar = 50  and   Sample size, n = 80

Let sample standard deviation be s.

96% confidence interval for the mean number of texts,\mu is given by ;

    96% confidence interval for \mu = xbar \pm 2.0537*\frac{s}{\sqrt{n} }

                              [46 , 54]           = 50 \pm 2.0537*\frac{s}{\sqrt{80} }

Since lower bound of confidence interval = 46

So,  46 = 50 - 2.0537*\frac{s}{\sqrt{80} }

       50 - 46 = 2.0537*\frac{s}{\sqrt{80} }

            s = \frac{4*\sqrt{80} }{2.0537} = 17.421

Therefore, standard deviation of the sample is 17.421 .

3 0
4 years ago
What is the value of the expression<br> 9r-5s+3 to the 3rd power if r=2/3 and s= -2
kramer

Answer:

43

Step-by-step explanation:

(9r-5s+3^3)

Let r = 2/3  and s = -2

(9 * 2/3-5* -2+3^3)

(6+ 10+27)

43

5 0
3 years ago
Read 2 more answers
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