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PIT_PIT [208]
3 years ago
6

You and your friend get braces on the same day. You go for a checkup every fourth Tuesday. Your friend goes every third Tuesday.

After how many weeks will you both be at the dentist on the same Tuesday?
Mathematics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer: After 12 weeks they will both be at the dentists on the same Tuesday

Step-by-step explanation:

Since we have given that

He goes for a checkup every fourth Tuesday and his friend goes for a check up every third Tuesday.

We need to find the number of weeks they both will be at the dentist on the same Tuesday.

For this we will find "L.C.M. of 4 and 3 " =12

So, After 12 weeks they will both be at the dentists on the same Tuesday.

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You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your p
frutty [35]

Answer:

Step-by-step explanation:

weight (x)                         ( x - \overline x)^2

12.05                               0.00123

12                                     0.00019

11.96                                0.00292

12.14                                0.0159

12.12                                0.00112

11.97                                0.0019

11.96                               0.0029

12.11                                 0.0092

11.94                                 0.0055

11.82                                0.038

12                                     0.00019

12.18                                0.028

11.91                                  0.0108

12.06                                0.0021

11.99                                  0.0029

\sum x = 180.21                    \sum (x - \overline x)^2 = 0.12285

Sample size = 15

\overline x = \dfrac{180.21}{15} = 12.014

Sample standard deviation

x = \sqrt{\dfrac{0.12285}{15-1}}

x = 0.0937

\text{degree of freedom = n - 1}

degree of freedom = 15- 1=14

\text{confidence interval} \alpha = 1- 0.98 = 0.02 \\ \\ \alpha/2 = 0.02/2 = 0.01

From \  X^2  \  table}\text{, the critical value} X^2 \text{=0.99 for degree of freedom =14 is given by : 4.660}

\text{, the critical value} X^2 \text{=0.01 for degree of freedom =14 is given by : 29.141}

\text{Thus, the lower limit = 4.660, the upper limit = 29.141}

\text{the confidence interval of the standard deviation} \ \sigma \  is:

\sqrt{\dfrac{(n -1)s^2}{X^2_{\dfrac{\sigma}{2}}} } \le \sigma \le \sqrt{\dfrac{(n -1)s^2}{X^2_{1-\dfrac{\sigma}{2}}} }

replacing our values:

\sqrt{\dfrac{(15 -1)0.0937^2}{29.141} } < \sigma

= 0.06495 < \sigma < 0.1624

\mathbf{Thus; \  \sigma (0.06495 , 0.1624)}

3 0
3 years ago
Area of the Shaded Region rectangle within reactanlge
Alik [6]

Answer:

132

Step-by-step explanation:

First you need to find the area of both regions (shaded and not shaded)

To find the shaded regions area multiply length by width

20 x 8 = 160

Then find the area of the unshaded region

7 x 4 = 28

Finally subtract the unshaded region from the shaded region

160 - 28 = 132

7 0
3 years ago
The scale size is 2cm:4m
MaRussiya [10]
I'm not sure which one is the scale size so i'll solve for both...

Ratio is 2cm:4m
0.02m:4m or 2cm:400cm also work as ratios

400cm/2cm=200 (so the scale is 200X smaller than the original)

Option 1::

If the scale of the item is:

l=12cm and w=6cm

then we can solve for the actual size by multiplying the values by 200

l=12cm*200=2400cm or 24m

w=6cm*200=1200cm or 12m

Option 2::

If the original item has a length of 12m and a width of 6m then:

l=12m and w=6m

We can find the scale length by dividing by 200

l=12m/200=.06m or 6cm

w=6m/200=0.03m or 3cm
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Blizzard [7]

Answer: HOPE THIS HELPSS!!! please mark me brainliestt

11 by 20

             110

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3 years ago
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Commemorative coins come in packs of 15, while coin holders come in packs of 25. What are the least numbers of packs you should
Bond [772]
Ur answer will be 25
8 0
3 years ago
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