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3 years ago

6

In a class of 28 students, 19 have a cat and 6 have a dog. There are 5 students who do not have a cat or a dog. What is the prob

ability that a student who has a cat also has a dog?

2 answers:

VikaD [51]3 years ago

6 0

25/28 would be your answer

77julia77 [94]3 years ago

3 0

Answer:

19/28juice wrld is not gone he is sippn juice in a better wrld

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An experiment was conducted to compare the wearing qualities of three types of paint (A, B, and C) when subjected to the abrasiv

MaRussiya [10]

Answer:

A) H0: mean of A= mean of B= mean of C

H1: all means are unequal

B) 3.48194

C) 0.04515

D) H0 is accepted. There is significant difference among mean hours after which visibile abrasion is observed for each type of paint

Step-by-step explanation:

A) Here Anova test will be used as we are comparing three groups

B) mean of A= 229.6, SD of A= 158.1962, SE of A= 50.026

mean of B= 309.9 , SD of B= 147.8742, SE of B= 46.7619

mean of C= 427.8, SD of C= 196.8179, SE of C= 62.2393

overall mean= 322.43

degrees of freedom between groups= 3-1=2

degrees of freeddom of error= total number of observations - number of groups= 30-3= 27

total degrees of freedom= total number of samples-1= 30-1 =29

Sum of squares between groups=∑ sample size of group×(individual group mean- over all mean)²

=198772.4667

mean of sum of squares between groups= sum of squares between groups/ (number of groups -1)

= 198772.4667/2

sum of squares of error=∑∑ (individual observation of group- group mean)²

= 770670.9223

mean sum of squares of erros= sum of squares of error/(total number of samples-number of groups)

= 770670.92/27

= 28543.38

F- stat= mean sum of squares between groups/ mean sum of squares of error

F stat= 3.48194

C) p- value= 0.04515

D) since F-stat is greater than p- value, null hypothesis is rejected

from F distribution table

F(2,27)=

3 0

3 years ago

Help please! the picture is below!

Thepotemich [5.8K]

The answer would be B

4 0

3 years ago

HELP ASAP THIS DUE TODAY FOR MY BRO

Harman [31]

Answer:

I think it's 10 or 10.5 not sure

Step-by-step explanation:

4 0

2 years ago

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).

Therefore, AB must be exactly half of AD:

$AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24$

3 0

3 years ago

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An artist painted a mural measuring 9’ x 20 1/2’ find the area and perimeter of the mural

Ann [662]

The difference is 180 is the correct answer for the problems

7 0

3 years ago