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2 years ago

Can anyone start me off on this? its trigonometric ratios

Mathematics

1 answer:

Alexxx [7]2 years ago

5 0

Answer:

The angle is about 30.27 degrees.

Step-by-step explanation:

Use the trig ratio cosine because you are given the adjacent leg and the hypotenuse.

$cos^{-1} =19/22$

Plug into a calculator

Rounded to the nearest hundredth the angle is 30.27 degrees.

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The average summer temperature in Anchorage is 69°F. If the daily temperature is normally distributed with a standard deviation

mafiozo [28]

Answer:

81.85%

Step-by-step explanation:

Given :

The average summer temperature in Anchorage is 69°F.

The daily temperature is normally distributed with a standard deviation of 7°F .

To Find:What percentage of the time would the temperature be between 55°F and 76°F?

Solution:

Mean = $\mu = 69$

Standard deviation = $\sigma = 7$

Formula : $z=\frac{x-\mu}{\sigma}$

Now At x = 55

$z=\frac{55-69}{7}$

$z=-2$

At x = 76

$z=\frac{76-69}{7}$

$z=1$

Now to find P(55<z<76)

P(2<z<-1)=P(z<2)-P(z>-1)

Using z table :

P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185

Now percentage of the time would the temperature be between 55°F and 76°F = $0.8185 \times 100 = 81.85\%$

Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.

7 0

3 years ago

1) stacey is selling tickets to the school play the tickets are $8 for adults and $5 for children . she sell twice as many ticke

Lapatulllka [165]

2) 25/48 All you do is multiply both.
1) she sold 5 children and 10 adult. 5 *5 equals 25. You double 5 to get twice as many tickets for adults and get 10. 8 *10 equals 80. 80 + 25= $105

5 0

3 years ago

A farmer has a basket of peaches. He gives ⅓ of the peaches to one person, ¼ to another, ⅕ to another, ⅛ to another, and then gi

inessss [21]

Answer:

$\frac{1429}{120}$ or $11\frac{109}{120}$

Step-by-step explanation:

Given:

A farmer has a basket of peaches. He gives ⅓ of the peaches to one person, ¼ to another, ⅕ to another, ⅛ to another, and then gives 7 peaches to a 5th person.

Remaining peaches = 4

We need to find the original number of peaches in the basket.

The farmer gives the total number of peaches = $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{8}+7$

Let x be the former gives the total number of peaches

We multiply and divide by 120 in right side of the above equation because of 120 is divided by all given denominator and then simplify.

$x = \frac{120}{120}(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{8}+7)$

$x = \frac{1}{120}(\frac{120}{3}+\frac{120}{4}+\frac{120}{5}+\frac{120}{8}+7\times 120)$

$x = \frac{1}{120}(40+30+24+15+840)$

$x = \frac{1}{120}(949)$

$x = \frac{949}{120}$

We add the remaining peaches by given peaches for the original number of peaches in the basket.

Original number of peaches = $\frac{949}{120}+4$

Original number of peaches = $\frac{949+4\times 120}{120}$

Original number of peaches = $\frac{949+480}{120}$

Original number of peaches = $\frac{1429}{120}$

Original number of peaches = $11\frac{109}{120}$

Therefore the original number of peaches in the basket is $\frac{1429}{120}$ or $11\frac{109}{120}$

5 0

3 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

3 years ago

I will give brainliest

natka813 [3]

Answer:

its 6

Step-by-step explanation:

5 0

3 years ago

Can anyone start me off on this? its trigonometric ratios​

Can anyone start me off on this? its trigonometric ratios