How do you solve proportions

PLEASE HELP FAST.....WILL MARK AS BRAINLIEST IF GIVEN GOOD ANSWER

Lady bird [3.3K]

Answer:

False, the dilation factor is 2.

Step-by-step explanation:

Notice how each measure of the segments for the red quadrilateral is 2x the amount of the segments for the blue triangle? That right there is the dilation factor.

5 0

3 years ago

What is the estimate of 90/7 divided by 1 3/4

attashe74 [19]

Answer:

7×17/49

Step-by-step explanation:

90/7 × 4/7

90×4 / 7×7

360/49

7 17/49

6 0

3 years ago

What is the value of x in the triangle?

Maksim231197 [3]

Answer:

5.619

Step-by-step explanation:

SOH 22 =x/15

cross multiply

X =SOH 22 *15

X =0.3746*15

X =5.619

6 0

3 years ago

diana has z gerbils. jackie has 4 times as many gerbils as diana. if w stands for the number of gerbils jackie has, express w in

disa [49]

D=z, j=4d,

If w=j, then w=4d,

If w=4d and d=z then:

w=4z

8 0

3 years ago

Clarinex is a drug used to treat asthma. In clinical tests of this drug, 1655 patients were treated with 5- mg doses of Clarinex

bagirrra123 [75]

Answer:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

$\hat p=0.021$ estimated proportion of interest

$p_o=0.012$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:

Null hypothesis: $p \leq 0.012$

Alternative hypothesis: $p > 0.012$

When we conduct a proportion test we need to use the z statisitic, 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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

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 $\alpha=0.01$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

3 0

3 years ago