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3 years ago

What is the product written in scientific notation?

1 answer:

8 0

$(1.18\cdot10^3)\cdot(9.1\cdot10^{-6})=(1.18\cdot9.1)(10^3\cdot10^{-6})\\\\=10.738\cdot10^{3+(-6)}=1.0738\cdot10\cdot10^{-3}\\\\=1.0738\cdot10^{1+(-3)}=1.0738\cdot10^{-2}\\\\Used:\\\\a^n\cdot a^m=a^{n+m}$

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Which two angles in the figure are obtuse angles?

ValentinkaMS [17]

Answer: C. ∠SQP and ∠TQR

Step-by-step explanation:

Since, by the given diagram,

m∠TQR = m∠SQP (vertical angles)

And, m∠TQP < m∠TQR

By adding ∠TQR on both sides,

m∠TQP + m∠TQR < m∠TQR+m∠TQR

⇒m∠TQP + m∠TQR < 2 ∠TQR

⇒ 180° < 2 m∠TQR ( Because ∠TQP and ∠TQR are linear angles)

⇒ 90° < m∠TQR

Hence ∠TQR is a obtuse angle,

⇒ ∠SQP also is a obtuse angle.

4 0

3 years ago

Write the fraction as a percent if necessary round to the nearest tenth of a percent 5 2/5

Olenka [21]

Answer:

$540\%$

Step-by-step explanation:

we have

$5\frac{2}{5}$

step 1

Convert the mixed number to an improper fraction first

$5\frac{2}{5}=\frac{5*5+2}{5}=\frac{27}{5}$

step 2

To convert the fraction as a percent multiply by 100

$\frac{27}{5}*100=27*20=540\%$

8 0

3 years ago

What is the largest number of identical sets that you can create by dividing the fish and plants evenly? Explain. 24 FISH 18 PLA

katovenus [111]

So you need one plant and one fish
there are 24 fish and 18 plants, you need one for one
18 plants and 18 fish

6 0

4 years ago

Read 2 more answers

In the basketboll gome yesterday. Brionna

Talja [164]

She would make 7 of her shots shooting or 7/20 percent which is 35% shooting. You can also multiply 20x0.35 and it will give you 7.

6 0

3 years ago

find the approximate probability that a randomly selected aspen tree in this park in 1975 would have a diameter less than 5.5 in

soldi70 [24.7K]

Answer:

The approximate probability that a randomly selected aspen tree in this park in 1975 would have a diameter less than 5.5 inches = 0.15866

Step-by-step explanation:

The complete question is presented in the attached image to this answer.

It is stated that the distribution of tree diameters is approximately normal, hence, this is a normal distribution problem with

Mean diameter = μ = 8 inches

Standard deviation = σ = 2.5 inches

The approximate probability that a randomly selected aspen tree in this park in 1975 would have a diameter less than 5.5 inches = P(x < 5.5)

To solve this, we first normalize or standardize 5.5 inches

The standardized score for 45mg/L is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (5.5 - 8)/2.5 = - 1.00

The required probability

P(x < 5.5) = P(z < -1.00)

We'll use data from the normal probability table for these probabilities

P(x < 5.5) = P(z < -1.00) = 0.15866

Hope this Helps!!!

5 0

4 years ago