Clea estimates that a glass contains 250.55 mL of water. The actual amount of water in the glass is 279.48 mL. HELP PLEASE

Answer:

98765432123459 that is te answer

Step-by-step explanation:

Answer:

1,2,5,10

Step-by-step explanation:

Greatest common factor is 10

Answer:

33

Step-by-step explanation:

tom will have his key 10

Answer:

40,320.. took the test on USA test prep

Step-by-step explanation:

Answer:

2.56

Step-by-step explanation:

8% of 32 is 2.56

Answer:

sorry if im wrong but im pretty sure -5/3

Step-by-step explanation:

Answer:

2/1

Step-by-step explanation:

A calculator showed me that this was the answer. I would give you the link but I would probably get deleted. I already got warned so yeah.

Hope this helps!

Answer:

-2

Step-by-step explanation:

It would be =

-2(-2 + 5) = -6

4 - 10 = -6

-10 and a positive four is a -6

Answer:

7

Step-by-step explanation:

So:a=3

b=(-2)

c=2

Substitute the numbers into each letters

so a is represented by 3,c by 2 and b -2

therefore giving us: 2

3+(-2)=9+(-2)=7

Answer:

The answer is C because you have to multiply them both:)

Answer:

The answer is C

Step-by-step explanation:

What I'm about to write is overkill for this question, but it works for any common denominator. Factor the denominators.

10: 2 * 5

5 : 1 * 5

You need 1 five and 1 two. The answer is 10

The correct way to go about doing the question is to multiply the 5 by 2 and the 2 on top by 2

2*2/5*2 = 4/10

1/10        1 / 10

4/10 + 1/10

Answer:

a

The 95% confidence interval is [tex] 0.2474 < p < 0.3188 [/tex]

b

  The result obtained does not contradict expectation

Step-by-step explanation:

From the question we are told that

   The number of green peas is  k =  438

   The number of yellow peas is u =  173

Generally the sample size is mathematically represented as

       [tex]n = k + u[/tex]

=>    [tex]n = 438 + 173[/tex]

=>    [tex]n = 611[/tex]

Generally the sample proportion for yellow peas is

      [tex]\^ p = \frac{u}{n}[/tex]

=>   [tex]\^ p = \frac{173}{611 }[/tex]

=>   [tex]\^ p = 0.2831[/tex]

From the question we are told the confidence level is  95% , hence the level of significance is    

      [tex]\alpha = (100 - 95 ) \%[/tex]

=>   [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally the margin of error is mathematically represented as  

     [tex]E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} } [/tex]

=>  [tex]E =  1.96 * \sqrt{\frac{ 0.2831 (1- 0.2831)}{611} } [/tex]

=>  [tex]E =  0.0357 [/tex]

Generally 95% confidence interval is mathematically represented as  

      [tex]\^ p -E <  p <  \^ p +E[/tex]

=>     [tex] 0.2831  -0.0357 <p< 0.2831  + 0.0357[/tex]

=>     [tex] 0.2474 < p < 0.3188 [/tex]

From the question we are told that it was expected that 25% of the offspring peas will be yellow

   Now from the 95% confidence interval obtained we see that the expected sample proportion(25% ) falls within it so  it means that the result obtained does not contradict expectation

Answer:

22

Step-by-step explanation:

Answer:0.27855153

A

−

0.04642525

kilograms per dollar

Step-by-step explanation:

Answer:

-2

Step-by-step explanation:

base on my stuck knowledge

Answer:

27

Step-by-step explanation:

6/10=12/20 because when you cross multiply, 120=120, so then you can do wither 6/10=x/45 and when you cross multiply you get 270=10x and 270/10=27.

Answer:

(A) The probability that a bulb burns between 778 and 834 hours is 0.51118.

(B) The probability that the average light bulb life is at least 820 hours in a random sample of 40 bulbs  is 0.00079

Step-by-step explanation:

Let X denote the life time of light bulbs.

It is provided that X follows a normal distribution with mean of 800 hours and standard deviation of 40 hours.

(A)

Compute the probability that a bulb burns between 778 and 834 hours as follows:

[tex]P(778<X<834)=P(\frac{778-800}{40}<\frac{X-\mu}{\sigma}<\frac{834-800}{40})\\\\=P(-0.55<Z<0.85)\\\\=P(Z<0.85)-P(Z<-0.55)\\\\=0.80234-0.29116\\\\=0.51118[/tex]

Thus, the probability that a bulb burns between 778 and 834 hours is 0.51118.

(B)

Compute the probability that the average light bulb life is at least 820 hours in a random sample of 40 bulbs as follows:

[tex]P(\bar X\geq 820)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}\geq \frac{820-800}{40/\sqrt{40}})\\\\=P(Z>3.16)\\\\=1-P(Z<3.16)\\\\=1-0.99921\\\\=0.00079[/tex]

Thus, the probability that the average light bulb life is at least 820 hours in a random sample of 40 bulbs  is 0.00079

Using the normal distribution and the central limit theorem, it is found that there is a:

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

Item a:

The probability is the p-value of Z when X = 834 subtracted by the p-value of Z when X = 778, hence:

X = 834:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{834 - 800}{40}[/tex]

[tex]Z = 0.85[/tex]

[tex]Z = 0.85[/tex] has a p-value of 0.8023.

X = 778:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{778 - 800}{40}[/tex]

[tex]Z = -0.55[/tex]

[tex]Z = -0.55[/tex] has a p-value of 0.2912.

0.8023 - 0.2912 = 0.5111

0.5111 = 51.11% probability that a bulb burns between 778 and 834 hours.

Item b:

Sample of 40, hence [tex]n = 40, s = \frac{40}{\sqrt{40}}[/tex].

The probability is 1 subtracted by the p-value of Z when X = 820, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{820 - 800}{\frac{40}{\sqrt{40}}}[/tex]

[tex]Z = 3.16[/tex]

[tex]Z = 3.16[/tex] has a p-value of 0.9992.

1 - 0.9992 = 0.0008

0.0008 = 0.08% probability that the average light bulb life is at least 820 hours.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213

Answer:

1.9 degrees above the normal temperture

Step-by-step explanation:

Normal temperture: 98.6 degrees Fahrenheit

100.5 - 98.6 =  1.9  

Answer:

36

Step-by-step explanation:

Median is the Middle number in the sequence.

Hope this Helps!!

:D

y=0.625x+1.75 is the equation for your problem.

Answer:

A.

Step-by-step explanation:

That is a translation of 3 units left and 6 units down.

Answer:

31.4

Step-by-step explanation:

You subtract 80.00 from 48.60, and you get your answer 31.4

Answer:

128.6

Step-by-step explanation:

You have to do the inverse operation.

So you add 48.60 to each side of the equation. This cancels out the -48.60 which isolates the variable. And on the other side of the equation it leaves you with 128.6. Therefore d = 128.6

Answer:

y = 3x + 10

Step-by-step explanation:

Answer:

im pretty sure it is (0,5) or (10,0)

Step-by-step explanation:

Answer:

6 degrees Celsius

Step-by-step explanation:

It started at 2 degrees Celsius below zero. The next day, it dropped to 12 degrees Celsius below zero. The next day, it rose to 7 degrees below zero. The next day, it increased to 6 degrees ABOVE zero. -7+13=6

Answer:

x= 3d/k +8b/k i think

Step-by-step explanation:

Answer:

0.8809

Step-by-step explanation:

Given that:

The population proportion p = 4% = 4/100 = 0.04

Sample mean x = 16

The sample size n = 300

The sample proportion [tex]\hat p =\dfrac{x}{n}[/tex]

= 16/300

= 0.0533

∴

[tex]P(\hat p \leq 0.0533) = P\bigg ( \dfrac{\hat p - p}{\sqrt{\dfrac{P(1-P)}{n}}}\leq\dfrac{0.0533 - 0.04}{\sqrt{\dfrac{0.04(1-0.04)}{300}}}\bigg )[/tex]

[tex]P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{\sqrt{\dfrac{0.04(0.96)}{300}}}\bigg )[/tex]

[tex]P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{\sqrt{\dfrac{0.0384}{300}}}\bigg )[/tex]

[tex]P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{\sqrt{1.28 \times 10^{-4}}}\bigg )[/tex]

[tex]P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{0.0113}}\bigg )[/tex]

[tex]P(\hat p \leq 0.0533) = P\bigg ( Z\leq1.18}\bigg )[/tex]

From the z tables;

= 0.8809

OR

Let X be the random variation that follows a normal distribution;

Then;

population mean [tex]\mu[/tex] = n × p

population mean [tex]\mu[/tex] = 300 × 0.04

population mean [tex]\mu[/tex] = 12

The standard deviation [tex]\sigma = \sqrt{np(1-p)}[/tex]

The standard deviation  [tex]\sigma = \sqrt{300 \times 0.04(1-0.04)}[/tex]

The standard deviation [tex]\sigma = \sqrt{11.52}[/tex]

The standard deviation  [tex]\sigma = 3.39[/tex]

The z -score can be computed as:

[tex]z = \dfrac{x - \mu}{\sigma}[/tex]

[tex]z = \dfrac{16 -12}{3.39}[/tex]

[tex]z = \dfrac{4}{3.39}[/tex]

z = 1.18

The required probability is:

P(X ≤ 10) = Pr (z  ≤ 1.18)

= 0.8809

Answer:

58.6 + 17.8 + x = 180

Step-by-step explanation:

hope this helps

Answer:

Me too not trying to get the points but how are u suppose to solve this anyway

Step-by-step explanation:

Answer:

Following are the solution to the given point:

Step-by-step explanation:

Please find the comp[lete question in the attached file.

Given:

[tex]\bold{ \lim_{n \to \ \infty} (1+ \frac{r}{n})^{nt} =e^{rt}}[/tex]

In point 1:

[tex]\to y = (1+ \frac{r}{n})^{nt}[/tex]

In point 2:

[tex]\to \ln (y)= nt \ln(1+ \frac{r}{n})[/tex]

In point 3:

Its key thing to understand, which would be that you consider the limit n to[tex]\infty[/tex],  in which r and t were constants!  

[tex]=lim_{n \to \ \infty} \ln (y) = lim_{n \to \ \infty} nt \ln(1+\frac{r}{n})\\\\= lim_{n \to \ \infty} \frac{\ln(1+\frac{r}{n})}{\frac{1}{nt}}\\\\= lim_{n \to \ \infty} \frac{\frac{-r}{\frac{n^2}{(1+\frac{r}{n})}}}{- \frac{1}{n^2t}}\\\\= lim_{n \to \ \infty} \frac{\frac{rn^2t}{n^2}}{(1+\frac{r}{n})}\\\\= lim_{n \to \ \infty} \frac{rt}{(1+\frac{r}{n})}\\\\= \frac{rt}{(1+\frac{r}{0})}\\\\=rt[/tex]

In point 4:

[tex]\to \lim_{n \to \ \infty} = (1+\frac{r}{n})^{nt}[/tex] and

[tex]\to \lim_{n \to \ \infty} = A_0e^{rt}[/tex]