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3 years ago

Write the words for: (3/4 - 1/5) x 4:

Mathematics

2 answers:

adoni [48]3 years ago

3 0

Answer:

three-forth minus one-fifth times four

Step-by-step explanation:

makkiz [27]3 years ago

3 0

Answer:

Step-by-step explanation:

Third -fourth subtracted from one-fifth multiplied by four

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myrzilka [38]

100 ounces = 6.25 pints

A mile unit is equal to 1760 yards. inches =mile * 63360 2*63360=126720

2 Miles = 126720 Inches

There are 36 inches in a yard

3600/36= 100

3600 inches = 100 yards

1 m = 1/1000 km

6800/1000 = 6.8

6800 m = 6.8 km

1 m = 100 cm

15000/100=150

15000 cm=150 m

a) Johnson ran yard in second= 9

1 minute = 60 second

Johnson ran yard in one minute= 9 *60= 540 yards

b) Johnson ran yard in second= 9

1 hour = 60 minutes *60 second= 1200 seconds

Johnson ran yard in one hour= 9 *1200= 10,800 yards

c) Johnson ran yard in one hour= 9 *1200= 10,800 yards

1 mile = 1760 yards

miles did he run in one hour= 10,800 / 1760 =6

Johnson ran 6 mile in one hour

5 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

2 years ago

sales taxes in the town where Amanda lives at 7% Amanda paid $35 in sales taxes on a new stereo what was the price of the stereo

Tema [17]

$37.45 hbf bgdfgfdbgfnhmjndbfvbdtjmuknhydbgfvck,iytnhjbgfvds

8 0

2 years ago

In a sequence described by a function, what does the notation f(3) = 1 mean?

LuckyWell [14K]

Answer:

The third term in the sequence has a value of 1.

7 0

3 years ago

Jimmy loaded three bags of sand into his wheelbarrow. The bags weighed 24

Rudik [331]

The whole numbers add up to 100. So now we are left with 1/2 or 0.5, 3/5 or .60, and 1/8 or .125. Those add up to 1.225 so the amount of sand in pounds in the wheelbarrow is 101.225. Hope I helped! If you have any questions just leave a comment.

3 0

3 years ago