All categories

3 years ago

PLZ HELP!!!!!!!!!!!!

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

6 0

Answer:

1. 0.5

2. 0.25

3. 0.33333...

4. 0.2

5. 0.75

6. 0.666...

7. 0.4

8. 0.375

9. 0.444...

10. 0.28571428571429

Step-by-step explanation:

You might be interested in

What is the total length of all the seeds<br> that the students measured?

mr_godi [17]

Answer:

525cmlong the length of seeds

8 0

3 years ago

I tried multiplying it but I still can’t figure it out please please help

neonofarm [45]

The answer is 88 5/6 . If you were stuck try turning the fractions into improper fractions and multiply both of them and later divide the numerator by the denominator to have a mixed number. You can later simplify the fraction to get 88 5/6. If you need help don't be afraid to tell me in the comments! Hope this helps :)

3 0

3 years ago

On a number line, a number, b, is located the same distance from 0 as another number, a, but in the opposite direction. The numb

vfiekz [6]

Answer:

Step-by-step explanation:

5 0

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

Help with part A and part B

Vitek1552 [10]

A. You can use y=mx+b to find the rate of change of a bushel corn in the current year. The rate of change would be y=8x.

B. One bushel is worth 8$ (24/3 = 8) in the current year, and in the previous year one bushel was worth 7$ (21/3= 7). Therefore there was a 1$ raise from the previous year to the current year.

8 0

3 years ago