Can anyone help me im in 6th grade i need answers

A ball is thrown upward with an initial velocity of 96 ft/sec from a height 640 ft. Its height h, in feet, after t seconds is

olchik [2.2K]

Check the picture below.

thus is at 0 = -16t² + 96t +640,

$\bf 0=-16t^2+96t+640\implies 0=-16(t^2-6t-40)
\\\\\\
0=t^2-6t-40\implies 0=(t-10)(t+4)\implies t=
\begin{cases}
\boxed{10}\\
-4
\end{cases}$

well, clearly it can't be a negative value for the elapsed seconds, so it can't be -4.

5 0

3 years ago

Cuanto es 4/5 de 55???????

Natalka [10]

Es 44. Esta página está en inglés, entonces nadie te va a responder la pregunta D=. Deberían de hacer una versión en español de esta página.

7 0

3 years ago

20. A rectangle has a length of x + 2 meters and a width of 2x - 5. (A.APR.1)

bija089 [108]

Answer:

A:

2(x + 2) + 2(2x - 5) =

2x + 4 + 4x - 10 =

6x - 6

perimeter = 6x - 6

B:

(x+2)(2x - 5) expand the brackets:

2x² - 5x + 4x - 10

area = 2x² - x - 10

C:

area = 6x - 6 = (6 x 5) - 6 = 30 - 6 = 24

perimeter = 2x² - x - 10 = (2 x 5²) - 5 - 10 = (2 x 25) - 5 - 10 = 50 - 5 - 10 = 35

brainliest please? x

6 0

3 years ago

The town librarian bought a combination of new-release movies on DVD for $20 and classic movies on DVD for $8. Let x represent t

IRISSAK [1]

answer: could be both (B)and(C)

Step-by-step explanation:

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

3 years ago

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