Given the angle 1=2x+35 and the angle 2=3x+7. Find the angle measures

Heheh ik im annoying but plssss im trying to graduate hs

zlopas [31]

Answer:

It's the 2nd one : -30 $a^{5}$ b + 12 $a^{4} b^{3}$ + 16 $a^{3} b^{2}$ - 4 $a^{2} b^{4}$ - 2a $b^{3}$

4 0

2 years ago

A large container holds 8 gallons of water. After 56 minutes the container only has 1 gallon of water left. At what rate is the

Korvikt [17]

Answer:

The rate at which water leaking is $\frac{1}{8}$ gal/min.

Step-by-step explanation:

Given:

A large container holds 8 gallons of water.

After 56 minutes the container only has 1 gallon of water left.

Now, to find at what rate is the water leaking.

So, water leaked from the container = 8 gallons - 1 gallon = 7 gallons.

Time taken for leakage = 56 minutes.

$\frac{Gallons\ of\ water\ leaked}{Time\ taken\ for\ leakage}$

$=\frac{7 gallons}{56 minutes}$

$=\frac{1 gallon}{8 minutes}$

The rate = $\frac{1}{8}$ gal/min.

Therefore, the rate at which water is leaking is $\frac{1}{8}$ gal/min.

4 0

2 years ago

Answer ASAP!

valentinak56 [21]

Answer:

A and D

Step-by-step explanation:

Since tangent is opposite/adjacent,

Tan 40 in this case would be x/3.8 (i used x because we don't know what the value is)

So, you set it up as an algebra problem

Tan40 = x/3.8

Multiply both sides by 3.8

3.8tan40 = x, Option A

And then, angle E is 50 degrees

So tan 50 = 3.8/x

Multiply both sides by x

tan50x = 3.8

Divide both sides by tan50

x= 3.8/tan50

So, A and D are both correct

3 0

2 years ago

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

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".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

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

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

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.