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Naily [24]
3 years ago
13

#4 find the value of x round answer to the nearest tenth

Mathematics
2 answers:
mel-nik [20]3 years ago
8 0

Answer:

x= 8.9

Step-by-step explanation:

Using our knowledge of trig functions

tan theta = opp/ adjacent

tan 25 = x /19

Multiply each side by 19

19 tan 25 = x

8.859845504944973263770117701191629574910197156776 =x

Rounding to the nearest tenth

8.9

AnnyKZ [126]3 years ago
6 0

Answer:

Step-by-step explanation:

Since this is a right angled triangle:

tan(∅) = opposite side/adjacent side

In this triangle:

tan(25°) = x/19

x = 8. 86 ≅ 8.9

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Step-by-step explanation:

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Graph the inequality <br> b-2&gt;-9
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Look at the picture i provided

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If N is the population of the colony and t is the time in​ days, express N as a function of t. Consider Upper N 0 is the origina
lesya [120]

Answer:

(a)N(t)=Noe^{kt}

(b)5,832 Mosquitoes

(c)5 days

Step-by-step explanation:

(a)Given an original amount N_o at t=0. The population of the colony with a growth rate k \neq 0, where k is a constant is given as:

N(t)=Noe^{kt}

(b)If N_o=1000 and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)

Therefore, our model is:

N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t

In 3 days time

N(3)=1000\cdot1.8^3=5832

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days

In approximately 5 days, the population of mosquitoes will be 20,000.

7 0
3 years ago
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This kind of stuff makes 0% sense to me
ser-zykov [4K]

Answer:

I think you're suppose to add the angles

Step-by-step explanation:

correct me if I'm wrong because as you can see ab CD are connected together even if you're still stuck I think the best thing is to ask your teacher.

5 0
3 years ago
A distribution of scores for a test of life stressors has a mean of µ = 125 and standard deviation of σ = 15. The researcher cal
laila [671]

Answer:

The mean and standard deviation for the z-scores in this distribution are 0 and 1 respectively.

Step-by-step explanation:

Let the random variable <em>X</em> follow a Normal distribution with mean <em>μ </em>and standard deviation <em>σ.</em>

The <em>z</em>-scores are standardized form of the raw scores <em>X</em>. It is computed by subtracting the mean (<em>μ</em>) from the raw score <em>x</em> and dividing the result by the standard deviation (<em>σ</em>).

z=\frac{x-\mu}{\sigma}

These <em>z</em>-scores also follow a normal distribution.

The mean is:

E(z)=E[\frac{x-\mu}{\sigma} ]=\frac{1}{\sigma}\times [E(x)-\mu] =\frac{1}{\sigma}\times [\mu-\mu]=0

The standard deviation is:

Var(z)=Var[\frac{x-\mu}{\sigma} ]=\frac{1}{\sigma^{2}}\times [Var(x)-Var(\mu)] =\frac{\sigma^{2}-0}{\sigma^{2}}=1\\SD(z)=\sqrt{Var(z)}=1

Thus, the mean and standard deviation for the z-scores in this distribution are 0 and 1 respectively.

8 0
3 years ago
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