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3 years ago

Help please I really need help, thank you

Mathematics

1 answer:

nirvana33 [79]3 years ago

6 0

Answer:

What do you need help with? Answer and I will tell you anything you need.

Step-by-step explanation:

You might be interested in

Rhonda had $300 to buy supplies for the animal shelter. She bought 8 bags of dog food and 11 leashes. Each bag of food cost $25.

Readme [11.4K]

Answer:

47.33

Step-by-step explanation:

Hey there! I'm happy to help!

We have 8 bags of dog food and each costs $25.19

8(25.19)=201.52

And we have 11 leashes each costing $4.65.

11(4.65)=51.15

We add these totals up.

201.52+51.15=252.67

We subtract this from 300 to see how much is left.

300-252.67=47.33

Rhonda had $47.33 left after her purchases.

Have a wonderful day! :D

3 0

3 years ago

3x + 3 = x - 5<br> Solve for x

Mars2501 [29]

Answer:

x = -4

Step-by-step explanation:

plz vote brainliest :)

4 0

3 years ago

Read 2 more answers

URGENT. Find f(3) if<br> f(x)=x^3 + 2x^2-x-1

KonstantinChe [14]

Answer:

Step-by-step explanation:

Jus replace x with 3 and put it in a calculator.

6 0

3 years ago

Read 2 more answers

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

Two ropes are attached to a large ballon in a parade. The people holding the ropes are 40 feet apart,and the angle between the t

qaws [65]

Ropes are 9.63 meters ans 8.178 meters.

<u>Step-by-step explanation:</u>

Step 1 :

List all the given things

Distance between people holding ropes = 40 =(40*0.3048) meters = 12.192 meters

Angle between two ropes = 86 degrees

Angle of elevation of one rope = 42 degrees

Step 2 :