How to do question five

What does the digit 6 represent in ?

marshall27 [118]

Answer:

6 is in ten place and its place value is 60, 3 is in ones place and its place value is 3.

4 0

3 years ago

Suppose that 12% of test scores for the unit 1 test were above 85 and 8% were below 70. Assuming a normal

Doss [256]

Answer:

Mean = 78.2

Standard deviation = 5.8

Step-by-step explanation:

Mathematically z-score;

= (x-mean)/SD

From the question;

12% of test scores were above 85

Thus;

P( x > 85) = 12%

P(x > 85) = 0.12

Now let’s get the z-score that has a probability of 0.12

This can be obtained from the standard normal distribution table and it is = 1.175

Thus;

1.175 = (85 - mean)/SD

let’s call the mean a and the SD b

1.175 = (85-a)/b

1.175b = 85 - a

a = 85 - 1.175b ••••••••(i)

Secondly 8% of scores were below 70

Let’s find the z-score corresponding to this proportion;

We use the standard normal distribution table as usual;

P( x < 70) = 0.08

z-score = -1.405

Thus;

-1.405 =( 70-a)/b

-1.405b = 70-a

a = 70 + 1.405b ••••••(ii)

Equate the two a

70 + 1.405b = 85 - 1.175b

85 -70 = 1.405b + 1.175b

15 = 2.58b

b = 15/2.58

b = 5.81

a = 70 + 1.405b

a = 70 + 1.405(5.81)

a = 78.16

So mean = 78.2 and Standard deviation is 5.8

4 0

3 years ago

find the digit in units place in the expansion of the following: 1) 2¹⁹⁶⁷ 2) 3¹⁹⁶⁹ 3)4¹⁹⁷⁰ 4) 4¹⁹⁷¹ 5) 5⁵⁵⁵ 6) 6¹⁰⁰⁰ 7) 7²⁰⁰⁷ 8)

diamong [38]

Answer:

troll

Step-by-step explanation:

troll!!!!!!!!!!!!!!!!!!!!!!

5 0

2 years ago

¿Una función racional es el<br>cociente irreducible de un polinomio?<br>Falso o Verdadero

Aliun [14]

es muy falso por qué no es un conciente irreductible

4 0

3 years ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

3 years ago