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3 years ago

Its about geometry i need help its urgent

Mathematics

2 answers:

Mnenie [13.5K]3 years ago

4 0

Answer:

1) 37

2) 23

3) x = 4

4) 58

Step-by-step explanation:

1) 15 + 22 = 37

2) 54 - 31 = 23

3) 6x + 1 + x + 7 = 36

7x + 8 = 36

7x = 28

x = 4

4) 3x + 10 + 47 = 9x - 39

57 = 6x -39

96 = 6x

16 = x

3(16) + 10 = 58

snow_tiger [21]3 years ago

3 0

Answer:

see below

Step-by-step explanation:

22 15

1. L -----------------M------------N LN = 22 + 15 LN = 37

31 ??

2. L -----------------M------------N MN = 54 - 31 MN = 23

|<-----------54 ---------------->|

6X+1 X+7

3. R -----------------S------------T find X

(6x + 1) + (x + 7) = 0

7x = -8

x = - 8/7

47 3X+10

4. D -----------E---------------------F find EF

|<--------- 9X - 39 ------------>|

47 + (3x + 10) = (9x - 39)

3x - 9x = -39 - 10 -47

-6x = -96

x = -96 / -6

x = 16

EF = 3x + 10

EF = 3(16) + 10

EF = 58

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3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

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

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

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

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

2 inches by 1.5 inches. The scale of the copy is one in.:40cm.

Nana76 [90]

Sorry I do not know. Sorry to let you down

5 0

3 years ago