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2 years ago

I NEED HELP ASAP!!!!!

Mathematics

1 answer:

lara31 [8.8K]2 years ago

3 0

Answer:

W = 2.5d + 62

Step-by-step explanation:

The calf weighed 62 pounds when they were born. This gives us a base of 62 pounds - the calf cannot weigh less than 62 pounds and it does on day 0. On each day , the calf gains 2.5 pounds. We can times the number to days by 2.5 to get the gain from day 0. We can add these two values together to get the total current weight of the calf.

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Multiply the trinomial x^2+xy+y^2 times the binomial x-y and simplify.

Natalija [7]

Multiply then simplify, that's doing the same work twice...anyway

x^3+x^2y+xy^2-x^2y-xy^2-y^3

x^3-y^3 which if factored is:

(x-y)(x^2+xy+y^2) which is what you started with :)

6 0

2 years ago

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the

Sergio039 [100]

Answer:

The estimate is $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is $\r p__{hat}} = 0.37$

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

$\alpha = 100 - 90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=> $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=> $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

$P__{hat}} \pm E$

substituting values

$0.37 \pm 0.0348$

5 0

2 years ago

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elena-s [515]

Answer:

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