Which of these problems would require regrouping before solving?

PLEASE HELP I NEED THIS ASAP<br> Find the vertex of y=x^2-7x+4

Goshia [24]

Answer:

$\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

Step-by-step explanation:

$y=x^2-7x+4\\\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}\\\mathrm{The\:parabola\:params\:are:}\\a=1,\:b=-7,\:c=4\\x_v=-\frac{b}{2a}\\x_v=-\frac{\left(-7\right)}{2\cdot \:1}\\\mathrm{Simplify}\\x_v=\frac{7}{2}\\\mathrm{Plug\:in}\:\:x_v=\frac{7}{2}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4\\$

$\mathrm{Simplify\:}\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4:\quad -\frac{33}{4}\\y_v=-\frac{33}{4}\\Therefore\:the\:parabola\:vertex\:is\\\left(\frac{7}{2},\:-\frac{33}{4}\right)\\\mathrm{If}\:a0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}\\a=1\\\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

6 0

3 years ago

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work i

USPshnik [31]

Answer:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 55, \pi = \frac{24}{55} = 0.4364$

93% confidence level

So $\alpha = 0.07$ , z is the value of Z that has a pvalue of $1 - \frac{0.07}{2} = 0.965$ , so $Z = 1.81$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 - 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.3154$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 + 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.5574$

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

8 0

2 years ago

This is 10 points plz help ( NO LINKS )

Natalija [7]

It’s D
16/4 is 4 no more no less

6 0

2 years ago

Read 2 more answers

The number of days in a month is measured in ones

krok68 [10]

The synodic month, or complete cycle of phases of the Moon as seen from Earth, averages 29.530588 mean solar days in length (i.e., 29 days 12 hours 44 minutes 3 seconds); because of perturbations in the Moon’s orbit, the lengths of all astronomical months vary slightly. The sidereal month is the time needed for the Moon to return to the same place against the background of the stars, 27.321661 days (i.e., 27 days 7 hours 43 minutes 12 seconds)

4 0

2 years ago

A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At

kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

$\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

(a) So, the probability of B,

$P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

$\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}$

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

$\implies P(A\cap B\cap C)=\frac{1}{8}$

7 0

2 years ago