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3 years ago

9

Solve for x!!!!!!!!! Spam will be taken down

1 answer:

IgorC [24]3 years ago

6 0

Answer:

10 10 $\sqrt{2}$

4 8

16 16 $\sqrt{3}$

$\sqrt{18}$ $\sqrt{18}$

11 22

384 192

Step-by-step explanation:

mark me brainliest pls

they are in order

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What are the like terms in this exspression 9a-3b + 4ab

vova2212 [387]

Answer:

none of these terms are like terms

7 0

3 years ago

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Two ratios by entering less than greater than or equal than 16 to 15, 30 to 80

Mumz [18]

Answer:

6:15>30:80

Step-by-step explanation:

The ratio;

6:15 written in the form x/1 is equivalent to 2/5 or 0.4

30:80 written in the form x/1 is equal to 3/8 or 0.375

0.4 is greater than 0.375

Therefore;

<u>6:15>30:80</u>

6 0

3 years ago

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

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

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

So

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

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

4 years ago

Can someone help out?

Tamiku [17]

Answer:

152.9 ft

Step-by-step explanation:

14*8=112

8*4=32

112+32+8.9=152.9

8 0

3 years ago

Can someone help me with this? thx :)

GrogVix [38]

Answer:

Table 1 is not a function because there are two y-values with the same x-values. Remember... y values can repeat but x values cannot repeat when their corresponding y values are different.

Table 2 is a function.

You would need to omit either (3,7) or (3,-2).

Please vote brainliest! Hope this helps, have a nice day.

8 0

3 years ago

Read 2 more answers