All categories

3 years ago

Which of the following is the maximum value of the function y = -x2 + 2x + 1?

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

We first complete the square.

y=-x^2+2x+1

y=-x^2+2x-1+1+1

y=-(x-1)^2+2

The maximum point is (1,2) since -1 in ( ) meant 1 unit to the right and +2 meant 2 units up.

The maximum point is (1,2) but the maximum value means the y value which is simply just 2.

Done!

You might be interested in

Alicia drives her car 287.5 miles in 5 hours. What is the constant of proportionality that relates the total number of miles, y,

Sphinxa [80]

Answer:

57.5

Step-by-step explanation:

The constant of proportionality is "k" in the equation ...

y = kx

It can be found from values of x and y using ...

k = y/x

For the given values, it is ...

k = (287.5 mi)/(5 h) = 57.5 mi/h

The constant of proportionality relating miles to hours is 57.5.

5 0

4 years ago

Need help with these math questions

PSYCHO15rus [73]

$\sqrt{16-x}$ x = 8

Since x = 8, you can plug in/substitute 8 for "x" in the equation:

$\sqrt{16-x}$

$\sqrt{16-8}$

$\sqrt{8}$

2.82842

2.83 You answer is the 4th option

$\sqrt{x+7}$ x = 9

Plug in 9 for "x" in the equation

$\sqrt{x+7}$

$\sqrt{9+7}$

$\sqrt{16}$

4 Your answer is the 1st option

4 0

3 years ago

SALE

vlada-n [284]

She would pay $132 for the chandelier

7 0

3 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZSelect all of the choices that are equal to (-5) - (-12).

valentina_108 [34]

Answer:

2 and 3

Step-by-step explanation:

subtract (-5) - (-12) and get 7 plus if you see a - that is finding the difference. therefore your answers are 2 and 3. Hope this helps! ^w^

4 0

3 years ago

In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion

S_A_V [24]

Answer:

99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Step-by-step explanation:

We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.

Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.

Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1+_n__2-2$

where, $\bar X_1$ = average gold thickness of control-immersion tip plating = 1.5 μm

$\bar X_2$ = average gold thickness of total immersion plating = 1.0 μm

$s_1$ = sample standard deviation of control-immersion tip plating = 0.25 μm

$s_2$ = sample standard deviation of total immersion plating = 0.15 μm

$n_1$ = sample of printed circuit edge connectors plated with control-immersion tip plating = 7

$n_2$ = sample of connectors plated with total immersion plating = 5

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2} }{7+5-2} }$ = 0.216

<em>Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

So, 99% confidence interval for the difference between the mean population mean, ( $\mu_1-\mu_2$ ) is ;

P(-3.169 < $t_1_0$ < 3.169) = 0.99 {As the critical value of t at 10 degree of

freedom are -3.169 & 3.169 with P = 0.5%}

P(-3.169 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 3.169) = 0.99

P( $-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.99

P( $(\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.99

<u>99% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ]

= [ $(1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }$ , $(1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }$ ]

= [0.099 μm , 0.901 μm]

Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

6 0

4 years ago

Read 2 more answers