All categories

3 years ago

Can pls someone help me plsss I need help :(

Mathematics

1 answer:

liraira [26]3 years ago

4 0

Answer: $7,5.66,39^{\circ}$

Step-by-step explanation:

Given

$IJ=9\\\angle I=51^{\circ}$

from figure, we can write

$\Rightarrow \sin 51^{\circ}=\dfrac{HJ}{9}\\\\\Rightarrow HJ=9\sin 51^{\circ}=6.99\approx 7$

$\Rightarrow \cos 51^{\circ}=\dfrac{HI}{9}\\\Rightarrow HI=9\cos 51^{\circ}\\\Rightarrow HI=5.66$

In right angle trianle, the sum of the remaining two angles is $90^{\circ}$

$\therefore 51^{\circ}+\angle J=90^{\circ}\\\Rightarrow \angle J=90^{\circ}-51^{\circ}\\\Rightarrow \angle J=39^{\circ}$

You might be interested in

Write the following number in standard decimal form four tenths

Crazy boy [7]

0.4 is four tenths. 0.4 = 4/10=2/5

8 0

3 years ago

Read 2 more answers

A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t

11Alexandr11 [23.1K]

Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

$dm/m=0.2dt$

b) Integrate

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) Antilogarithm

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

Use the measured rate of 4 grams per hour after 3 hours

$t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

Particular solution:

$m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

3 0

3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago

Lena found a 5/8

DENIUS [597]

Answer:

Naomi had the correct measurement the ring weighed 0.625 carats

Step-by-step explanation:

* The diamond ring weighed 5/8 carats

- To know who is right lets change the fraction 5/8 to a decimal number

∵ 5/8 means 5 ÷ 8

- 5 is smaller than 8 so we will multiply it by 10 and

insert decimal point in the quotient (answer of division)

∴ It will be 50 ÷ 8 = 6 and remainder 2/8 ⇒ 1/4

∴ The quotient = 0.6 and remainder 1/4

- 1 is smaller than 4 so we will multiply it by 10

∴ It will be 10 ÷ 4 = 2 and remainder 2/4 ⇒ 1/2

∴ The quotient = 0.62 and remainder 1/2

- 1 is smaller than 2 so we will multiply it by 10

∴ It will be 10 ÷ 2 = 5 without remainder

∴ The quotient = 0.625

* Naomi had the correct measurement the ring weighed 0.625 carats

8 0

3 years ago

Read 2 more answers

Please answer this correctly Is the answer 6:07am 9:11am 6:39am 10:11am

nika2105 [10]

Answer:10:11 am

Step-by-step explanation:

3 0

3 years ago