Three times a number w is less than 18. The inequality is

Please help me I’m so lost !!!!

Tasya [4]

Answer:

$4095$

Step-by-step explanation:

$\sum _{n=1}^63\left(4^{n-1}\right)\:$

$\sum _{n=1}^63\cdot \:4^{n-1}$

Compute general progression:

$\frac{3\cdot \:4^{\left(n+1\right)-1}}{3\cdot \:4^{n-1}}=4$

$r=4\\$

$a_1=3\cdot \:4^{1-1}$

$a_1=3$

nth term is computed by:

$r=4,\:a_n=3\cdot \:4^{n-1}$

plug in the values $n=6,\:\spacea_1=3,\:\spacer=4$ :

$=3\cdot \frac{1-4^6}{1-4}$

$=4095$

5 0

3 years ago

Theresa is planning on making a fleece blanket for her nephew. She learns that the fabric she wants to use is $7.99 per yard. Fi

Anastaziya [24]

Answer:

$31.96

Step-by-step explanation:

(7.99)×(4)= $31.96

7 0

3 years ago

Read 2 more answers

I need help please! I've been stuck on this for a week and its really stressful lol

worty [1.4K]

Answer:

$y=0.2*4^{x}$

Step-by-step explanation:

Notice when x increases 1, y is 4 times the previous one, so

the function is like $y=C*4^{x}$

To determine the constant C, put any pair of (x, y)

Use x = 0, y = 0.2, so

0.2 = $C*4^{0}$ = C * 1 = C

then $y=0.2*4^{x}$

5 0

2 years ago

Read 2 more answers

Assignment The operation teeter is defined on the set T= { 2, 3, 5, 7 } by x teeter y=[x+y+xy) mod.8 a - Construct mod.8 table f

rosijanka [135]

Answer:

Case A) tau_net = -243.36 N m, case B) tau_net = 783.36 N / m, tau_net = -63.36 N m, case C) tau _net = - 963.36 N m,

Explanation:

For this exercise we use Newton's relation for rotation

Σ τ = I α

In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N

we will assume that the counterclockwise turns are positive.

case a

tau_net = m g x - F x2

tau_nett = -28.8 9.8 1.5 + 180 1

tau_net = -243.36 N m

in this case the man's force is downward and the system rotates clockwise

case b

2 force clockwise, the direction of

the force is up

tau_nett = -28.8 9.8 1.5 - 180 2

tau_net = 783.36 N / m

in case the force is applied upwards

3) counterclockwise

tau_nett = -28.8 9.8 1.5 + 180 2

tau_net = -63.36 N m

system rotates clockwise

case c

2 schedule

tau_nett = -28.8 9.8 1.5 - 180 3

tau _net = - 963.36 N m

3 counterclockwise

tau_nett = -28.8 9.8 1.5 + 180 3

tau_net = 116.64 Nm

the sitam rotated counterclockwise

4 0

1 year ago

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

3 years ago