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olya-2409 [2.1K]
3 years ago
7

3. The sides of one triangle are 6, 8, and 10. The sides of another triangle are 10, 24, and

Mathematics
1 answer:
irakobra [83]3 years ago
3 0

Answer: i think its either 3:5 or 1:3 or there just not similar

Step-by-step explanation:

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A model of a rectangular patio at a landscaping business will be enlarged by a scale factor of 2 when it is installed in a custo
VARVARA [1.3K]

Answer:

C)  The area of the landscape model is A = 40 sq ft.

Step-by-step explanation:

The original dimensions of the rectangular patio model  is

Length = L

Width = W

Area of the patio model  = LENGTH x WIDTH =   L  x W

⇒ A  =  L W .............  (1)

Now, the new area A" is enlarged by a factor of 2

⇒  The new Length = L"  = (2 L)

     The new Width  = W"  = (2 W)

So, AREA"  = L" x W" = (2 L) x (2 W)  = 4 (L W)

⇒  A "  =    4 (L W)

But, L W = A   .. from (1)

⇒   A"  = 4  A

But,  the area of the new enlarged patio is 160 square feet.

⇒   160 sq ft  = 4 x A

or, A   = 160 / 4 =40 sq ft

⇒ A = 40 sq ft.

Hence, the area of the landscape model is A = 40 sq ft.

5 0
4 years ago
Read 2 more answers
Show that cos3(B)+cos(B)sin2(B) = 1 sin(B)cos2(B)+sin3(B) tan(B)
Pachacha [2.7K]

Answer:

-90

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A countertop is 16 feet long and 3 feet wide. what is the area of the countertop in square meters
Paladinen [302]

The area of the countertop is (16-ft x 3-ft) = 48 feet² .

1 foot = 0.3048 meter (rounded)

(1 foot)² = (0.3048 meter)² = 0.0929 meter²

48 feet² = (48 x 0.0929) =  4.459 meter² (rounded)

8 0
3 years ago
A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po
igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b) Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c)z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation    

X_{1}=170 represent the number of people with the characteristic 1

X_{2}=110 represent the number of people with the characteristic 2  

n_{1}=200 sample 1 selected  

n_{2}=150 sample 2 selected  

p_{1}=\frac{170}{200}=0.85 represent the proportion estimated for the sample 1  

p_{2}=\frac{110}{150}=0.733 represent the proportion estimated for the sample 2  

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

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

\alpha=0.05 significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

a.State the decision rule.

For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c. Compute the value of the test statistic.                                                                                              

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

5 0
3 years ago
Last week, Jada ran 2 7/10 miles. This week, she plans to run 5 times as far as last week.
V125BC [204]

Answer:

13.5 miles or 27/5 miles

Step-by-step explanation:

2 7/10 is equal to 2.7  If she plans to run 5 times as much as last week, multiply 2.7 by 5.

2.7*5=13.5

5 0
4 years ago
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