Mandy was flying in a helicopter at an altitude of 500 feet. The helicopter descended 450 feet. What is Mandy's altitude now

The measured length is 250 ft. The actual length is 246.9 ft. Find the percent error.

Marta_Voda [28]

Percent error = 1.2556%

Steps:
Percent Error =
Vobserved - Vtrue
Vtrue
=
250 - 246.9
246.9
=
3.1
246.9
= 1.2555690562981%

7 0

3 years ago

What are the steps to(2x+3y)4

lilavasa [31]

(2x + 3y)4

Distribute

4*2x = 8x
4*3y = 12y

8x + 12y

Answer: 8x + 12y

8 0

3 years ago

Read 2 more answers

CAN SOMEONE PLEASE HELP ME!!!!

love history [14]

Answer:

The balance after the payment is $1263.84.

Step-by-step explanation:

The formula for amount after compound interest is

$A=P(1+\frac{r}{n})^{t}$

Where, P is principal, r is rate of interest, n is number of time interest compounded in a period, number of periods.

According to the given information,

P=1455.69

r=0.128

n=365

t=45

Put these values in the above formula,

$A=1455.69(1+\frac{0.128}{365})^{45}$

$A\approx 1478.84$

The amount after compound interest is $1478.84. Add late fee chages $35 in this amount and subtract the payment of $250. So, the balance amount after payment is

$Balance=1478.84+35-250=1263.84$

Therefore the balance after the payment is $1263.84.

8 0

3 years ago

The first element in an arithmetic sequence is 2. Its twenty second is 14. Find the value of n so that a(n)=6

Aleks04 [339]

Answer:

Step-by-step explanation:

Given the nth term of an arithmetic sequence to be Tn = a+(n-1)d

a = first term of the sequence

n = number of terms

d = common difference.

Given the first element a = 2 and 22nd to be 14

T22 = a+(22-1)d = 14

a+21d = 14

Substtuting a = 2 into the equation to get d

2+21d = 14

21d = 12

d = 12/21

d = 4/7

The nth term of the sequence given a = 2 and d = 4/7 will be expressed as;

Tn = 2+(n-1)4/7

Given Tn = 6

6 = 2+(n-1)4/7

6 = 2+4/7 n - 4/7

6-2+4/7 = 4/7 n

32/7=4/7 n

32 = 4n

n = 32/4

n = 8

8 0

3 years 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