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castortr0y [4]
1 year ago
13

Arrange the following numbers in order from greatest to least 7,361; 7,136; 7,613

Mathematics
2 answers:
d1i1m1o1n [39]1 year ago
3 0
7,613 ,7,361,7,136

 hope that helped
Zanzabum1 year ago
3 0
7,613, 7,361, 7136

Hope this helps <3
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The store sells a 9 ounce jar of mustard for $1.53 and a 15 ounce for $2.55 explain whether the cost of the mustards have the sa
astraxan [27]

Answer:

same amount of money

Step-by-step explanation:

3 0
1 year ago
Match the following items.
Margarita [4]

Answer:

1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40

Step-by-step explanation:

Given:

\angle DBC=40°

From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.

m\textrm{ arc CD}=2\times \angle DBC\\m\textrm{ arc CD}=2\times 40=80

Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,

m\angle BCD=90

From the Δ BCD, which is a right angled triangle,

m\angle DBC+m\angle BDC=90\textrm{ (right angled triangle)}\\40+m\angle BDC=90\\m\angle BDC=90-40=50

Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.

Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles m\angle DBC\textrm{ and }m\angle BDC and EF is a tangent to the circle at point C.

Therefore, m\angle DCF=m\angle DBC=40\\m\angle ECB=m\angle BDC=50

Again, using the same theorem as above,

m\angle DCF=50\\\therefore m\textrm{ arc BC}=2\times m\angle DCF=2\times 50=100

Hence, all the angles are as follows:

1. m\angle ECB=50\\2. m\textrm{ arc BC}=100\\3. m\textrm{ arc CD}=80\\4. m\angle DCF=40

6 0
1 year ago
Read 2 more answers
What number would you divide by to calculate the mean of 3, 4, 5, and 6?
Mademuasel [1]

Answer:

4

Step-by-step explanation:

Mean = sum of scores /number of scores

We have four scores given ,so, the sum of the scores given will be divided by the number of scores which is 4

4 0
2 years ago
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The average weight of a chicken egg is 2.25 ounces with a standard deviation of 0.2 ounces. You take a random sample of a dozen
n200080 [17]

Answer:

P(\bar X

P(\bar X

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sigma)  

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(2.25,\frac{0.2}{\sqrt{12}})

Solution to the problem

We want this probability:

P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X

P(\bar X

5 0
2 years ago
A box in a certain supply room contains four 40w, five 60w, and six 75w light-bulbs. suppose that three bulbs are randomly selec
weeeeeb [17]
Situation satisfies the criteria for the use of hypergeometric distribution. Since no replacement is made, binomial distribution is not applicable (probability does not remain constant).

A=number of target wattage bulbs
B=number of non-targeted wattage bulbs
a=number of target wattage bulbs selected
b=number of non-targeted wattage bulbs selected

P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)
where C(n,x)=combination of x items chosen from n=n!/(x!(n-x)!)

For all following problems, 
A+B=4+5+6=15
a+b=3 (selected)

(a) Target wattage = 75W
A=6, B=9, a=2, b=1
P(a,b,A,B)
=P(2,1,6,9)
=C(6,2)*C(9,1)/C(15,3)
=15*9/455
=27/91

(b) target wattage = each of the three
Probability = sum of probabilities of choosing 3 40,60,75-watt bulbs
P(3x40W)+P(3x60W)+P(3x75W)
Case    (A,B,a,b)
3x40W (4,11,3,0)
3x60W (5,10,3,0)
3x75W(6,9,3,0)

P(3x40W)+P(3x60W)+P(3x75W)
=C(4,3)*C(11,0)/C(15,3)+C(5,3)*C(10,0)/C(15,3)+C(6,3)*C(9,0)/C(15,3)
=4*1/455+10*1/455+20*1/455
=34/455

Can also be solved by elementary counting, for example, for (a),
P(2x75W)
=C(3,2)*6/15*5/14*9/13
=(3)*6/15*5/14*9/13
=27/91 as before
8 0
1 year ago
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