If m 2 BOC = 27 and m ZAOC = 61, what is the measure of ZAOB?

Help with thissss please thx

Monica [59]

The perimeter is 47cm. The area would be 111cm^2. Hope this helped, could I possibly get brainliest?

5 0

3 years ago

Evaluate. Write your answer as a whole number or as a simplified fraction.<br> 12-².4³ =<br> Submit

Delicious77 [7]

$12^{-2} \times 4^3\\\\=(4 \times 3)^{-2} \times 4^3\\\\=4^{-2} \times 4^3 \times 3^{-2}\\\\=4^{-2+3} \times 3^{-2}\\\\=4^1 \times \dfrac1{3^2}\\\\=\dfrac 49$

7 0

2 years ago

-20y + 15 = 2 - 16y + 11

shtirl [24]

In order to answer this, we will first need to combine like terms on each side. on the left, you can leave them alone. however on the right, we will need to combine 2 and 11. this is 13. the right side becomes 13-16y. after that, we can add 20y to both sides. that equals 15=13+4y now we can subtract 13 from both sides. 2=4y. then we divide by 4 on both sides to find y. y=.5

7 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

Subject is Math pls help look at the photo :)

Stels [109]

1. 50
2. 154
3. 28
These would be your answers

5 0

2 years ago