All categories

2 years ago

I NEED HELP ON 7, 9, AND 10! PLEASE

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

4 0

✧･ﾟ: *✧･ﾟ:* Hello ✧･ﾟ: *✧･ﾟ:*

༶•┈┈⛧┈♛ ♛┈⛧┈┈•༶

I will give you a explanation about how I got the answer!

Plus, I'm am so sorry if I answer this so long...

Hope It Helped! :)

Read more on simple interest on:

brainly.com/question/24429956

3 0

2 years ago

The librarian noticed that 60% of seventh graders checked out fantacy books. About how many of 240 seventh graders would check o

ehidna [41]

Answer:

The correct answer is 144.

Step-by-step explanation:

The librarian noticed that 60% of total seventh graders checked out fantasy books from the library.

Total number of seventh graders is given to be 240.

Number of seventh grade students who checked out fantasy books = 60% of 240 = 0.6 × 240 = 144.

Thus out of 240 seventh graders 144 of them would check out a fantasy book from the library.

4 0

3 years ago

A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

3 years ago

092:15 PMEach month, henry weeds gardens for 15 hours and mows lawns for x hours.For each job, he is paid $9 per hour. He wants

ankoles [38]

We know that he weeds gardens for 15 hours, and mows lawns for x hours.

If he wants to earn $252 per month, we can express the following

$9\cdot15+9x\ge252$

Now, we solve for x

$\begin{gathered} 135+9x\ge252 \\ 9x\ge252-135 \\ 9x\ge117 \\ x\ge\frac{117}{9} \\ x\ge13 \end{gathered}$ <h2>Hence, he has to mow lawns for 13 hours to earn at least $252.</h2>

6 0

1 year ago