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Oduvanchick [21]

3 years ago

5

Willa bought 1 yard of yarn to knit a scarf. She wants to cut the yarn into for equal parts. Draw and label a number line from 0

to 1 to show where Willa would cut her yard of yarn. Label all the fractions.

1 answer:

SSSSS [86.1K]3 years ago

4 0

Answer:

Step-by-step explanation:

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C(x)=4x-2 and d(x)=x^(2)+5, what is (c*d)(x)

lawyer [7]

Answer:4x^3-2x^2+20x-10

Step-by-step explanation:

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

*<br> HELP: Find the indicated angle measure.<br> Find m2NLM<br> A. 10<br> B.60<br> C.120<br> D.30

Debora [2.8K]

The answer is easy
So we need to fund m2NLM
The answer is c

5 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago

Anyone know the answer please teach me

Semenov [28]

Answer: Look at step by step

Step-by-step explanation:

(a) There are 119 stocks that increased in price out of 246 stocks so the probability of selecting a stock that is increased in price is 119/246 = 48.37%

(b) There are 112 stocks where it pays dividends so the probability so the probability of selecting a stock that pays dividends is 112/246 = 45.53%

(c) There are 49 stocks which satisfy the question so the probability is 49/246 = 19.92%

(e) There are 34 stocks which increase in price and pay dividends out of the 119 stocks that increase in price so the probability is 34/119 = 28.57%

(f) Using the same logic as (e), the probability is 85/134 = 63.43%

(g) From the requirements, we see that 49 stocks don't satisfy the requirements, so the probability is (246-49)/246 = 80.08%

7 0

1 year ago