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pshichka [43]
3 years ago
12

What is the value of y when multiple by b

Mathematics
1 answer:
7nadin3 [17]3 years ago
6 0

What?? I am very sorry, but this is confusing.

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The two figures are similar. Find the ratios (red to blue) of the perimeters and of the areas. Write the ratios as fractions in
vesna_86 [32]

Answer:

see explanation

Step-by-step explanation:

The perimeters of the similar figures have the same ratio as the sides.

ratio of perimeters = \frac{11}{6}

ratio of areas = 11² : 6² , that is 121 : 36

ratio of areas = \frac{121}{36}

3 0
3 years ago
Round to the nearest thousand. 5,568
cricket20 [7]

Answer:

Round to the nearest thousand.

5,568

= 5,000

7 0
3 years ago
Read 2 more answers
Please help and thank you so much in advanced
sdas [7]

x=18

y=-6

To find the x intercept, you would substitute the y with 0 and then solve for x.

To find the y intercept, you would substitute the x with 0 and solve for y.

Hope this helps braniliest please!

8 0
3 years ago
Read 2 more answers
SOMEONE PLEASE HELP ME I WILLG IVE EVERYTHING!
Anvisha [2.4K]

Answer:

<h2>AB is around 33.18</h2><h2>BC is around 15.58 </h2>

Step-by-step explanation:

adjacent/hypotenuse is sine:

cos(28 degrees)=29.3/x

cos(28 degrees)*x=29.3

x=29.3/cos(28 degrees)

x=around 33.18

AB is around 33.18

opposite/adjacent is tangent

tan(28 degrees)=x/29.3

tan(28 degrees)*29.3=x

x=tan(28 degrees)*29.3

x=around 15.58

BC is around 15.58

4 0
3 years ago
4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

8 0
3 years ago
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