Which expression is equivalent to 18

Use the frequency table to determine how many students received a score of 80 or better on an English exam.

Sveta_85 [38]

10 students scored above an 80

3 0

3 years ago

Mi mamá Rosa Amelia, asistió al Hospital cercano a su casa, luego de hacer atendida, el medico tratante le dio tres tipos de pas

kirill [66]

Answer:

La próxima pastilla roja será a las 12:00 a.m.

La próxima pastilla amarilla será a las 8:00 p.m.

La próxima pastilla blanca será a las 7:00 p.m.

4 0

2 years ago

Suppose that we want to test the hypothesis with a significance level (alpha) of .05 that the climate has changed since industri

saw5 [17]

Answer: we reject the initial claim

(50. 3802, 51.6198)

Step-by-step explanation:

A)

The initial statement is that the mean temperature throughout history is 50°

This is the null hypothesis and it is denoted as H'

H': u = 50.

After taking a sample of 40 years, we realized that the mean temperature is 51°.

This is the alternative hypothesis, in contradiction to the alternative.

The alternative hypothesis is denoted as H1

H1: u > 50 ( upper tailed).

Sample size (n) = 40

Sample mean (x) = 51

Population standard deviation (σ) = 2

We use a z test to get the value of the test statistics.

We are using a z test because sample size is greater than 30 ( n = 40) and population standard deviation is given.

Z score = x - u/ (σ/√n)

Z score = 51 - 50 / (2/√40)

Z score = 1 / 0.3162

Z score = 3.16

Our level of significance is 5%, and the critical value at this level of significance is 1.645.

By comparing the z score relative to the critical value, since our z score is greater than 1.645, it implies that we are in the rejection region hence we reject the initial claim.

B)

We are to construct a 95% confidence interval for population mean temperature.

For upper limit

u = x + Zα/2 ×(σ/√n)

Where Zα/2 = 1.96 which is the critical value for a two tailed test at 5% level of significance.

For upper limit, we have that

u = 51 + 1.96 × (2/√40)

u = 51 + 1.96 (0.3162)

u = 51 + 0.6198

u = 51.6198.

For lower limit, we have that

u = 51 - 1.96 × (2/√40)

u = 51 - 1.96 (0.3162)

u = 51 - 0.6198

u = 50. 3802.

Hence the 95% confidence level for mean temperature is given as (50. 3802°, 51.6198°)

7 0

2 years ago

Logx - logx-1^2=2log(x-1)

myrzilka [38]

Answer:

x is approximately 2.220744

Step-by-step explanation:

This can be simplified a little using properties of logarithms, and then solve it by graphing:

$log(x)-log(x-1)^2=2\,log(x-1)\\log(x)-2\,log(x-1)=2\,log(x-1)\\log(x)=4\,log(x-1)$

So we use a graphing tool to find the intersection point of the graph of $log(x)$ , and the graph of $4\,log(x-1)$

Please see attached image for the graph and solution.

The value of x is approximately 2.220744

8 0

3 years ago

Read 2 more answers

Plz guys help i need asap<br> its part b im stuck

aleksklad [387]

Answer:

a.2nd quarter with 9 goals

b. 4.8 goals

c. 4 goals

Step-by-step explanation:

a. The mode is defined as the most appearing data point or the data point with the highest frequency..

From our data(for away goals):

1st quarter-2
2nd quarter-9
3rd quarter-7
4th quarter-4

Hence, the 2nd quarter has the mode for away goals with 9 goals.

b. Mean is defined as the average of a set of data points.

#We calculate the totals goals per quarter, sum over all quarters then divide by the number of games, 10:

$\bar x=\frac{1}{n}\sum{x_i}\\1^{st }_g=Away+Home=5+2=7\\\\2^{nd}_g=Away+Home=4+9=13\\\\3^{rd}_g=Away+Home=8+7=15\\\\4^{th}_g=Away+Home=9+4=13\\\\\bar x=\frac{1}{n}\sum{x_i}=\frac{1}{10}(7+13+15+13)=4.8$

Hence, the mean number of goals per quarter is 4.8 goals

c. To find the number of more home goals than away goals, we subtract from their summations as:

$g_m=\sum{g_h}-\sum{g_a}\\\\=(5+4+8+9)-(2+9+7+4)\\\\=26-22\\\\=4$

Hence, there are 4 more home goals than away goals.

4 0

2 years ago

Which expression is equivalent to 18 - 7?