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3 years ago

If a figure is dilated by a scale factor less than one, does the shape get larger or smaller?

Mathematics

2 answers:

malfutka [58]3 years ago

6 0

I think that it gets smaller. (not entirely sure)

lesya692 [45]3 years ago

3 0

<span>If a figure is dilated by a scale factor less than one, the shape gets SMALLER.

</span>--------------------------------------------------------------------------
<span>But if the figure is dilated by a scale factor more than one, the shape gets bigger</span>

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If two sides of your triangle are 6cm and 18cm, what is the range of values that the third side could be?

Gnom [1K]

Answer:

The ranges of the value can be from 12-24

Step-by-step explanation:

Hope this helps <3

Tell me if you want an explanation.

5 0

3 years ago

Read 2 more answers

Nehal is a real estate developer, and he is designing a new neighborhood. The neighborhood will have four streets, each of which

ololo11 [35]

Answer:

240 houses

Step-by-step explanation:

Given that:

Number of streets = 4

Length of each street = 3/4 miles long

Street is divided into lots with one house built per lot

1 mile = 5289 feets

3/4 miles = (3/4) * 5280 = 3960 feets

Hence, street is 3960 feets long

Since each lot must have at least 65 feet frontage along the street:

Number of lots per street :

Length of street / frontage length

3960 ft / 65 ft = 60.92

Hence, maximum number of lots per street = 60 lots per street

Maximum number of houses in New neighborhoods :

Number of lots per street × number of streets

= 60 × 4

= 240 houses

4 0

3 years ago

WILL MARK YOU BRAINLIEST ASAP

34kurt

Answer:

(-0.5, -1)

Step-by-step explanation:

To find the midpoint add the coordinates then divide by 2

x = (2 + -1)/2

x = 1/2 = 0.5

y = (3 + -5)/2

y = -1

(-0.5, -1)

5 0

2 years ago

Read 2 more answers

The function f(x)= -6x+11 has a range given by {-37,-25,-13,-1}.Select the domain values of the function from the list 1,2,3,4,5

andreev551 [17]

The range is {-37,-25,-13,-1}. So you need to figure out what four numbers from this list of numbers (1,2,3,4,5,6,7,8), when applied to this
function, ( f(x)=-6x+11 ), equals these numbers that are in the range {-37,-25,-13,-1}.

So you apply each of these numbers (1,2,3,4,5,6,7,8) into the function (f(x)=-6x+11)
one by one.

f(1)=-6(1)+11=5
f(2)=-6(2)+11= -1
f(3)=-6(3)+11= -7
f(4)=-6(4)+11= -13
f(5)=-6(5)+11= -19
f(6)=-6(6)+11= -25
f(7)=-6(7)+11= -31
f(8)=-6(8)+11= -37
As you can see, f(2),f(4),f(6),and f(8) equal the numbers that are in the range {-37,-25,-13,-1}.

4 0

2 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