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2 years ago

Help pls pls help pls

Mathematics

1 answer:

galina1969 [7]2 years ago

6 0

Answer: E = 120.2

Step-by-step explanation:

just plug in the celsius for C

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Can somebody tell me how to solve these kind of problems?

Ugo [173]

Answer: 918km^2

Step-by-step explanation:

$area(a): 1700km^2\\percentage(p): 5.75 = \frac{5.75}{100}=0.0575\\years(y): 8$

Let x be the new dimension of the area;

$x=a-(a*p*y)$

$x=(1700km^2)-(1700km^2*0.0575*8)\\x=1700km^2-782km^2\\x=918km^2$

6 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

vaieri [72.5K]

x²(x - 4)(2x + 5) = 0

x² = 0 or x - 4 = 0 or 2x + 5 = 0

√x = √0 or x = 4 or 2x = -5

x = 0 x = $-\frac{5}{2}$

Answer: 0, 4, $-\frac{5}{2}$

8 0

3 years ago

Need help with dividing integer questions

lord [1]

8/2 is 4 and -10 + -2 is -12 because your adding two negatives

5 0

2 years ago

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Jennifer utilizo 2/7 de metro de cinta para hacer moños para sus primas. Ahora le quedan 10/21 de metro de cinta. ( problema ver

Vikki [24]

Answer:

$x = \frac{16}{21}\,m$

Step-by-step explanation:

El enunciado pregunta la cantidad inicial de cinta utilizada por Jennifer, la cual es la suma de la porción utilizada y la porción remanente:

$x = \frac{2}{7}\,m + \frac{10}{21}\,m$

$x = \frac{42}{147}\,m + \frac{70}{147}\,m$

$x = \frac{112}{147}\,m$

$x = \frac{16}{21}\,m$

6 0

2 years ago

40% of Oatypop cereal boxes contain a prize. Hannah plans to keep buying cereal until she gets a prize. What is the probability

aleksley [76]

Answer:

$(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

Step-by-step explanation:

We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.

Our success is winning the prize, p =40/100 = 0.4

Then failure q = 1-0.4 = 0.6

Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.

P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)

= $\binom{n}{0}p^{0}q^{n-0}$ + $\binom{n}{1}p^{1}q^{n-1}$ + $\binom{n}{2}p^{2}q^{n-2}$ + $\binom{n}{3}p^{3}q^{n-3}$

= $q^{n}+npq^{n-1}$ + $\frac{n(n-1)(p)^{2}q^{n-2}}{2}$ + $\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}$

= $(0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

7 0

2 years ago

Read 2 more answers