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Maurinko [17]
3 years ago
13

Rooster + rooster + rooster =20

Mathematics
1 answer:
Leya [2.2K]3 years ago
6 0
If rooster = 6 eggs =10 banana = 5 and 6+10x5= 56
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Cos X<br> Y<br> 12<br> X<br> 16<br> 20<br> Z<br> 4<br> A) 3/4<br> B) 4/3<br> C)3/5<br> D) 5/4
yanalaym [24]

Answer:

C.

\cos X =  \frac{12}{20}  \\  =  \frac{3}{5}

5 0
3 years ago
Jack needs to rent a car for his vacation. Car rental company A charges customers a one-time rental fee of $100 plus $0.10 for e
lapo4ka [179]

Answer:

I think it's B but I'm not sure

5 0
3 years ago
Read 2 more answers
The original price was $20,000 the sales price was $18,000 what is the percent discount
lord [1]

Answer:

10 percent :)

Step-by-step explanation:

Discount = Original Price x Discount %/100

Discount = 20000 × 10/100

Discount = 20000 x 0.1

You save = $2,000.00

Final Price = Original Price - Discount

Final Price = 20000 - 2000

Final Price = $18,000.00

i hope this helps! pls stay happy and healthy, friend! :))

3 0
3 years ago
Which expression is equivalent to 7h^2-252k^2?
gulaghasi [49]

Answer:

c

Step-by-step explanation:

252 = 7 *36

36 = 6 *6 = 6²

a² -b²  = (a + b)(a - b)

7h² - 252k² =  7(h² - 36k²)

                   = 7(h² - [6k]² )

                  = 7(h  +6k)(h - 6k)

6 0
3 years ago
Read 2 more answers
A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
professor190 [17]

Answer:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

n=15 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =16 represent the sample variance

\sigma^2_0 =25 represent the value that we want to  verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

The statistic is given by:

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

And replacing we got:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0
3 years ago
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