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2 years ago

Please help me please thank you

Mathematics

2 answers:

Gennadij [26K]2 years ago

8 0

Answer:

Step-by-step explanation:

a-c is a line

EastWind [94]2 years ago

8 0

Answer:

10. BJ

11. ABK

12. JBK

13. PRQ

14. PRC

15. KBR

Step-by-step explanation:

You might be interested in

There are 9 classes of 25 students each 4 teachers in two times as many chaperones as teachers each bus hold a total of 45 peopl

Tanya [424]

Answer:

Least number of bus require for trip = 5 buses (Approx)

Step-by-step explanation:

Given:

Total number of classes = 9

Number of student in each class = 25

Number of teacher = 4

Number of chaperones = Double of teacher

Bus hold = 45 people

Find:

Least number of bus require for trip

Computation:

Total number of student = 9 × 25

Total number of student = 225

Number of chaperones = 4 × 2

Number of chaperones = 8

Total people = 225 + 8 + 4

Total people = 237

Least number of bus require for trip = Total people / Bus hold

Least number of bus require for trip = 237 / 45

Least number of bus require for trip = 5.266

Least number of bus require for trip = 5 buses (Approx)

6 0

2 years ago

Consider the following system of equations:

Kobotan [32]

Answer:

x=-15, y=25. (-15, 25).

Step-by-step explanation:

2x+3y=45

x+y=10

--------------

x=10-y

2(10-y)+3y=45

20-2y+3y=45

20+y=45

y=45-20

y=25

x+25=10

x=10-25

x=-15

5 0

3 years ago

Read 2 more answers

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$