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yawa3891 [41]
3 years ago
8

All statements of equality below are correct. A.True B.False

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
5 0

Answer:

A

Step-by-step explanation:

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Help with math question
NNADVOKAT [17]

Well

You want to calculate the interest on $1000 at 3.5% interest per month after 3 year(s).

The formula we'll use for this is the simple interest formula, or:

Where: P is the principal amount, $1000.00.

r is the interest rate, 3.5% per month, or in decimal form, 3.5/100=0.035.

t is the time involved, 3....year(s) time periods.

Since your interest rate is "per month" and you gave your time interval in "year(s)" we need to convert your time interval into "month" as well.

Do this by multiplying your time, 3 year(s), by 12, since there's 12 months in 1 year.

So, t is 36....month time periods.

To find the simple interest, we multiply 1000 × 0.035 × 36 to get that:

The interest is: $1260.00

Hope i could help

6 0
3 years ago
∠abc measures 115° and ∠dbc measures 70°. What is the measurement of ∠abd?
olasank [31]

Answer:

45 °

Step-by-step explanation:

m< ABC = m< DBC + m< ABD

= 115 _ 70

= 45 °

3 0
3 years ago
with reference to a universal set, the inclusion of a subset in another is relation which is a) symmetric only b) transitive onl
Vlad [161]

Answer:

D?

Step-by-step explanation:

I think it is D since a universal set would only be {1,2,3,4,5,6,...} and the subset would not satisfy any of the conditions of symmetric, transitive, and reflexive.

5 0
3 years ago
A tire manufacturer warranties its tires to last at least 20,000 miles orâ "you get a new set ofâ tires." In itsâ experience, a
Y_Kistochka [10]

Answer:

Probability that a set of tires wears out before 20,000 miles is 0.1151.

Step-by-step explanation:

We are given that a tire manufacturer warranties its tires to last at least 20,000 miles or "you get a new set of tires." In its past experience, a set of these tires last on average 26,000 miles with S.D. 5,000 miles. Assume that the wear is normally distributed.

<em>Let X = wearing of tires</em>

So, X ~ N(\mu=26,000,\sigma^{2}=5,000^{2})

Now, the z score probability distribution is given by;

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

where, \mu = average lasting of tires = 26,000 miles

            \sigma = standard deviation = 5,000 miles

So, probability that a set of tires wears out before 20,000 miles is given by = P(X < 20,000 miles)

    P(X < 20,000) = P( \frac{X-\mu}{\sigma} < \frac{20,000-26,000}{5,000} ) = P(Z < -1.2) = 1 - P(Z \leq 1.2)

                                                                    = 1 - 0.88493 = 0.1151

Therefore, probability that a set of tires wears out before 20,000 miles is 0.1151.

4 0
3 years ago
Simplify (8 + 7i) + (2 –i)
inna [77]
<span>Simplify (8 + 7i) + (2 –i) 
</span>answer:10+6i 
8 0
3 years ago
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