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butalik [34]
3 years ago
7

One of the factors in a multiplication is less than 1. How does the product compare to the other factor?

Mathematics
1 answer:
xxMikexx [17]3 years ago
6 0
The product will be smaller than the other factor.
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Luke lives0.4 kilometer from the skating rink. Mark lives0.25 kilometer from the skating who lives closer
Alchen [17]

Answer:

mark

Step-by-step explanation:

Mark lives closer because 0.25 is less than 0.4

on the number line 0.25 is before 0.4

8 0
3 years ago
In the diagram abc=adb=90, ad=p and dc=q. Use similar triangles to show that x2=pz<br> plzz anyoneee
kramer

Answer:

By comparing the ratios of sides in similar triangles ΔABC and ΔADB,we can say that x^{2} =pz

Step-by-step explanation:

Given that ∠ABC=∠ADC, AD=p and DC=q.

Let us take compare Δ ABC and  Δ ADB in the attached file , ∠A is common in both triangles

                                                                     and given ∠ABC=∠ADB=90°

Hence using AA postulate, ΔABC ≈ ΔADB.

Now we will equate respective side ratios in both triangles.

\frac{AB}{AC}= \frac{AD}{AB}=\frac{BD}{BC}

Since we don't know BD , BC let us take first equality and plugin the variables given in respective sides.

\frac{x}{z}= \frac{p}{x}

Cross multiply

x^{2}=pz

Hence proved.


7 0
3 years ago
Pamela's age is two times Jiri's age. The sum of their ages is <br> 33<br> . What is Jiri's age?
notka56 [123]
Let x is Jiri's age
2x is Pamela's age

x + 2x = 33
3x =33
x = 33/3
x =11
Answer: Jiri's age is 11
4 0
3 years ago
Read 2 more answers
Can u help me please ?
harina [27]
Check the attached file and let me know what you think.
Download docx
6 0
3 years ago
Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars
GuDViN [60]

Answer:

Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

Step-by-step explanation:

We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.

<em>Let X = incomes for the industry</em>

So, X ~ N(\mu=95,\sigma^{2}=5^{2})

Now, the z score probability distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean income of firms in the industry = 95 million dollars

            \sigma = standard deviation = 5 million dollars

So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)

    P(X < 100) = P( \frac{X-\mu}{\sigma} < \frac{100-95}{5} ) = P(Z < 1) = 0.8413   {using z table]

                                                     

Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

5 0
3 years ago
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