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3 years ago

7.006 x 10^-3 in standard notation

Mathematics

1 answer:

frutty [35]3 years ago

7 0

Answer:

7.006*10⁻³ = 0.007006

Step-by-step explanation:

7.006*10⁻³ = 0.007006

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Which equation can be used to calculate the area of the shaded triangle in the figure below?

zzz [600]

Answer:

b. 1/2(12x4) = 24 square feet

Step-by-step explanation:

Using rectangle area formula and considering half of it as shaded:

A = 1/2bh

Given

b = 12 feet, h = 4 feet

The area is:

1/2*12*4 = 24 square feet

Correct option is b.

8 0

3 years ago

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The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the

garri49 [273]

Answer:

D. I and III only

Step-by-step explanation:

Number of first fie defects are given as 9,7,10,4 and 6.

First we have to arrange the above data in ascending order which is 4,6,7,9,10

Now if we consider the defect in sixth car to be 3 then our data will look like: 3,4,6,7,9,10

So the mean of above data would be $\frac{3+4+6+7+9+10}{6}$ = 6.5

and Median of the above data would be = $\frac{3^{rd} obs + 4^{th} obs}{2}$ = $\frac{6+7}{2}$ = 6.5

Hence mean and median number of defects are same if the sixth car has 3 defects.

Now if we consider the defect in sixth car to be 12 then our data will look like: 4,6,7,9,10,12

So the mean of above data would be $\frac{4+6+7+9+10+12}{6}$ = 8

and Median of the above data would be = $\frac{3^{rd} obs + 4^{th} obs}{2}$ = $\frac{7+9}{2}$ = 8

Hence mean and median number of defects are same if the sixth car has 12 defects.

But Now if we consider the defect in sixth car to be 7 then our data will look like: 4,6,7,7,9,10

So the mean of above data would be $\frac{4+6+7+7+9+10}{6}$ = 7.167

and Median of the above data would be = $\frac{3^{rd} obs + 4^{th} obs}{2}$ = 7

Hence mean and median number of defects are not same if the sixth car has 7 defects.

Therefore option D is correct.

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3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

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4 years ago

In 2005, 0.76% of all airline flights were on-time. If we choose a simple random sample of 2000 flights, find the probability th

marissa [1.9K]

Answer:

Step-by-step explanation:

Here X no of flight that reach on time has two outcomes only and each trial is independent of the other.

Since sample size is large we approximate to normal with mean =np and variance =npq

i.e. X is N(152,19.10)

$Z=\frac{x-152}{19.1}$ is used for converting to std normal variate

Continuity correction of 0.5 on either side of interval would be applied for x

a) $P(X\geq 79%) = P(X\geq 1580)\\=P(X\geq 1579.5)\\=P(Z\geq 3.12)$

=0.00

b) $P(X\leq 1580)\\=P(X\leq 1580.5)\\=P(Z\leq 3.17)\\=$ =1.00

c) p is N(0.76, sqrtpq/n)=N(0.76,0.0095)

P(p diff>0.03) =

$P(Z>\frac{0.03}{0.0095} \\=P(Z>3.16)$ =0.00

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4 years ago

Gary is playing a game using the spinner shown.Which statement about the spinner is true?

rosijanka [135]

Answer:

The probability of the pointer landing on yellow is equal to the probability of the pointer landing on red

5 0

3 years ago

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