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3 years ago

An educator believes that new reading activities for elementary school children will improve reading

Mathematics

1 answer:

Damm [24]3 years ago

6 0

The correct answer to this open question is the following.

Unfortunately, you forgot to attach the scores shown in the back-to-back stem-and-leaf display. So we do not know what the numbers are and we do not have any reference at all

What we can do to help you is to comment on the following general terms.

There have been previous and similar experiments or projects like this in other schools in America. These results suggest that the new activities are better because extra or special reading comprehension programs better prepare students to understand what they are reading and comprehend more than basic ideas of the text.

Students that participate in these programs develop a better sense to understand and like what they are reading, considerably increasing their focus and attention.

These programs have resulted positively when trying to improve the marks of the students, compared to other traditional approaches.

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The pizza costs £18.00. Mark agrees to pay 1/4 of the cost. Samuel agrees to pay 1/2 of the cost Imran agrees to pay the rest. H

MariettaO [177]

Answer:

4.5 or 4.50 dollars

Step-by-step explanation:

if we split 18.00 into $\frac{1}{4}$ each part of the fraction would be 4.5 . or 4.50

mark pays 4.50

samuel pays 9.00

add those both and you'll get 13.50 $

now subtract 13.50 from 18.00 ( og price)

imran pays 4 . 5 0

hope this helped :p

3 0

3 years ago

Help me please thanks

Brut [27]

Forget about the negative sign. First you have to find a least common denominator. This would be 63. Since we had to multiply 9 by 7 to get 63 we have to multiply the numerator (4) by 7. This gives us a new fraction of 36/63. Since we ha e to multiply 21 by 3 to get 63 we have to multiply 19 by 3. This gives us 57. So now we have 36/63 and 57/63. Then we add the negative sign back in making 57/63 -57/63. Now we add them together. Because one has a negative sign and one is positive the number has to be closer to zero. So the answer is -21/63.

3 0

3 years ago

On his first three tests, Hiram got scores of 89, 95, and 90. His final test score brought his median score to 91.5. What was I

Elena-2011 [213]

The median is the middle score, it is also the average of the two middle scores when total number is even.

original 3 scores: 89, 90, 95 ----> median = 90

with final score: 89,90,X,95 ----> median = 91.5

(90+x)/2 = 91.5
90+x = 183
x = 93

Final score was a 93.

6 0

3 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$