Find the quotient in simplified 8/5 divided by 3/4

Find the standard equation for the circle having the given center and radius.

RoseWind [281]

$\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{0}{h}~~,~~\underset{0}{k})}\qquad \stackrel{radius}{\underset{7}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 0 ~~ )^2 ~~ + ~~ ( ~~ y-0 ~~ )^2~~ = ~~7^2\implies x^2+y^2=49$

7 0

2 years ago

What pair of points have a slope of -3/5

sergey [27]

Answer:

(5,0) and (0,3) have a slope of -3/5

Step-by-step explanation:

Here, we want to state a pair of points that have a slope of -3/5

Mathematically, we can have the slope of a straight line calculated as;

m = y2-y1/(x2-x1)

This could give points like (5,0) and (0,3)

3 0

3 years ago

A researcher is interested in the lengths of Salvelinus fontinalis (brook trout), which are known to be approximately Normally d

kompoz [17]

Answer:

c. 0.1151

Step-by-step explanation:

d) the probability that, on average, fish are larger than 86 centimeters in length.The area under part of a normal probability curve is directly proportional to probability and the value is calculated as

z = (x₁−x) /σ

where z = propability of normal curve

x₁ = variate mean = 86cm

x = mean of 80cm

σ = standard deviation = 5cm

applying the formula,

z= (86-80)/5

z = 6/5 =1.2

Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 1.2 corresponds to an area of 0.3849 between the mean value. but, because the standard curve has 0.5, then will minus 0.3849 from 0.5= 0.5 - 0.3849 = 0.1151

Thus the probability of a fish are larger than 86 centimeters in length is 0.1151

5 0

4 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

There are four erasers but only three pencils write the ratio of erasers to pencils

iris [78.8K]

4:3
4 to 3
4/3
I wrote all the three ways

3 0

3 years ago