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3 years ago

What plane contains E,F,and H

Mathematics

2 answers:

Sati [7]3 years ago

7 0

...no one can answer u bc theres no pic soooo yea

kvasek [131]3 years ago

5 0

B. The plane on the bottom of the figure

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The diagonals of a square are 4 meters long. The side of this square is equal to the diagonal of a second square. Find the side

jok3333 [9.3K]

Check the picture below.

since the second figure is also a square, then the sides touching the diagonals have to be all equal, and that'd only happen if those sides bisects the larger square's diagonal.

3 0

3 years ago

Ian measured a city park and made a scale drawing. In real life, the soccer field is 114 meters long. It is 19 millimeters long

ioda

Answer:

1 mm : 6 m

Step-by-step explanation:

Take the 19 mm and divide it by 114 meters. to get the scale

19/114 = 1/6

1 mm : 6 meters

8 0

3 years ago

2/3x+2 -x-3 <br> solution and the steps how to do it

Stells [14]

$\text{Hey there!}$

$\mathsf{\dfrac{2}{3}x+2-x-3}\\\\\mathsf{COMBINE\ your\ like\ terms}\\\\\mathsf{(\dfrac{2}{3}x-x)+(2-3)}\\\\\mathsf{\dfrac{2}{3}x-x=\dfrac{-1}{3}}\\\\\mathsf{2-3=-1}\\\\\boxed{\boxed{Thus,\ your \ answer\ is: \boxed{\dfrac{-1}{3}x -1}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{LoveYourselfFIrst:)}$

7 0

3 years ago

Read 2 more answers

Suppose a shipment of 190 electronic components contains 3 defective components. To determine whether the shipment should be ac

PolarNik [594]

Answer:

0.0467

Step-by-step explanation:

Probabilty of being defective is $\frac{3}{190}$

So probability of being good is $\frac{187}{190}$

Now,

Probability that all 3 of them are good is:

$\frac{187}{190}^{3}=0.9533$

Now, to find probability of rejection, we subtract the now found probability from 1. Thus, we have:

1 - 0.9533 = 0.0467

5 0

3 years ago

A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain

svetlana [45]

Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

$\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55$

The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

$LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8$

The probability of having between 4 and 8 customers choosing the oversize version is:

$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

$P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328$

7 0

3 years ago