I’m A Triangle, The Sum Of Any Two Side Lenghts, Finish The Sentence

At Basketball practice Kyla made 60 free throws this was 80% of her attempts how many free throws did kyle attempt?

DanielleElmas [232]

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5 0

3 years ago

12x^2+20=50 what is x

LenaWriter [7]

Answer: x=1.5811389999999999 , or 1.59

Step-by-step explanation:

First, Subtract 20 from both sides.

12x2+20−20=50−20

12x2=30

Then, Divide both sides by 12:

12x^2/12 = 30/12

⇒ you get : x^2 = 5/2

After, Take square root.

x= ± √ 5/2

Finally you answer is going to be =1.5811389999999999 OR, 1.59

* Hopefully this helps:) Mark me the brainliest:)

∞ 234483279c20∞

7 0

3 years ago

Which binomial is a factor of 49x2 + 84 xy + 36y2?

bezimeni [28]

Answer:

(7x+6y) is a binomial factor of the given expression $49x^2+84xy+36y^2$

Therefore $49x^2+84xy+36y^2=(7x+6y)(7x+6y)$

Step-by-step explanation:

Given expression is $49x^2+84xy+36y^2$

To find the binomial factor :

$49x^2+84xy+36y^2$

Rewritting the above expression as

$=7^2x^2+2(7)(6)xy+6^2y^2$

$=(7x)^2+2(7x)(6y)+(6y)^2$

$=(7x+6y)^2$ ( using the formula $=(a+b)^2=a^2+2ab+b^2$ here a=7x and y=6y )

$=(7x+6y)(7x+6y)$ ( using the formula $=(a+b)^2=(a+b)(a+b)$ )

Therefore $49x^2+84xy+36y^2=(7x+6y)(7x+6y)$

(7x+6y) is a binomial factor of the given expression $49x^2+84xy+36y^2$

7 0

3 years ago

Anna saves $3 for every $10 she earns Patrick saves $5 for every $20 he earns do the two save money at the same rate

guapka [62]

No they are not at the same rate

4 0

3 years ago

A sample of 1000 college students at NC State University were randomly selected for a survey. Among the survey participants, 102

cupoosta [38]

Answer:

The upper endpoint of the 99% confidence interval for population proportion is 0.13.

Step-by-step explanation:

The confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Given:

n = 1000

$\hat p$ = 0.102

$z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58$

*Use the standard normal table for the critical value.

Compute the 99% confidence interval for population proportion as follows:

$CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.102\pm 2.58\times\sqrt{\frac{0.102(1-0.102)}{1000}}\\=0.102\pm0.0248\\=(0.0772, 0.1268)\\\approx (0.08, 0.13)$

Thus, the upper limit of the 99% confidence interval for population proportion is 0.13.

6 0

3 years ago