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prisoha [69]
3 years ago
14

In a litter of 8 puppies, suppose there are 2 brown females, 2 brown males, 1 spotted female, and 3 spotted males. If a puppy is

randomly selected, what is the probability of selecting a female or a brown puppy?
Mathematics
2 answers:
Alexandra [31]3 years ago
7 0
There is a total of eight puppies. there are two female and brown puppies, one additional female puppy, and two additional brown puppies. so the number of possible wanted outcomes is 5. the probability is 5/8 or 62.5 %
tigry1 [53]3 years ago
3 0

Answer:

spotted male

Step-by-step explanation:

if you look at the numbers then you will see that this one has the biggest number

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Using the Binomial Probability table, find P(X) when n = 7, X < 3, and p = 0.5 A. 0.273 B. 0.500 C. 0.227 D. 0.773
Nat2105 [25]

Answer:

0.227

Step-by-step explanation:

Given that X is binomial with n=7 and p =0.5

To find the probability for P(X<3)

i.e. we have to find prob as =P(x=0,1,2)

=P(X=0)+P(X=1)+P(x=2)

P(X=r)=7Cr(0.5)^7

Using binomial table we find prob

P(X<3)=0.2265

Round off this to 3 decimals as

0.227

Hence answer is 0.227

8 0
3 years ago
Select all that appear in the DOMAIN {(-8,-12),(4,-8), (2, -10),(-10.-16) -16 -12 -10 -8 -2-4
Yuri [45]
Domain of a set of ordered pairs

We know the domain is the set of all x when is represented by ordered pairs: (x, y)

In this case {(-8,-12),(4,-8), (2, -10),(-10.-16) } we can observe that there are four x (the first number of each pair):

Domain = { -8, 4, 2, -10}

<h2>Domain = {-10, -8, 2, 4}</h2>

3 0
1 year ago
Fill in the number that makes this sentence true 6x9=(6x5)+(6x)
antiseptic1488 [7]
<span>Fill in the number that makes this sentence true 6x9=(6x5)+(6x)           (6x5)</span>
4 0
2 years ago
A College Alcohol Study has interviewed random samples of students at four-year colleges. In the most recent study, 494 of 1000
Vinil7 [7]

Answer:

The 95% confidence interval for the difference between the proportion of women who drink alcohol and the proportion of men who drink alcohol is (-0.102, -0.014) or (-10.2%, -1.4%).

Step-by-step explanation:

We want to calculate the bounds of a 95% confidence interval of the difference between proportions.

For a 95% CI, the critical value for z is z=1.96.

The sample 1 (women), of size n1=1000 has a proportion of p1=0.494.

p_1=X_1/n_1=494/1000=0.494

The sample 2 (men), of size n2=1000 has a proportion of p2=0.552.

p_2=X_2/n_2=552/1000=0.552

The difference between proportions is (p1-p2)=-0.058.

p_d=p_1-p_2=0.494-0.552=-0.058

The pooled proportion, needed to calculate the standard error, is:

p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{494+552}{1000+1000}=\dfrac{1046}{2000}=0.523

The estimated standard error of the difference between means is computed using the formula:

s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.523*0.477}{1000}+\dfrac{0.523*0.477}{1000}}\\\\\\s_{p1-p2}=\sqrt{0.000249+0.000249}=\sqrt{0.000499}=0.022

Then, the margin of error is:

MOE=z \cdot s_{p1-p2}=1.96\cdot 0.022=0.0438

Then, the lower and upper bounds of the confidence interval are:

LL=(p_1-p_2)-z\cdot s_{p1-p2} = -0.058-0.0438=-0.102\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= -0.058+0.0438=-0.014

The 95% confidence interval for the difference between proportions is (-0.102, -0.014).

7 0
3 years ago
Consider M, N, and P. collinear points on MP.
Kobotan [32]

Answer:

See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}

2. Point N lies to the right from point P, then MP + PN = MN.  Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}

3 0
3 years ago
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