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Varvara68 [4.7K]

3 years ago

13

4p - 7 < 21

A/p < 4
B/p < 7
C/p > 7
D/p > 4 PLEASE HELP

1 answer:

Fofino [41]3 years ago

8 0

It would be e cause it needs to be

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Which set of whole numbers share the common multiples 2,520 and 3,780?

olya-2409 [2.1K]

The set of whole numbers that share the common multiples of 2,520 and 3,780 are: 35, 105, and 315.
To get these sets of whole numbers, you have to decompose 2520 and 3780 by using decomposition method or the continuous division.

See attached file.

7 0

3 years ago

Two angle measures of a quadrilateral are 34° and 66°. What could the measure of the other two angles be?

alexdok [17]

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3 years ago

Best known for its testing program, ACT, Inc., also compiles data on a variety of issues in education. In 2004 the company repor

GarryVolchara [31]

Answer:

$\mu_p -\sigma_p = 0.74-0.0219=0.718$

$\mu_p +\sigma_p = 0.74+0.0219=0.762$

68% of the rates are expected to be betwen 0.718 and 0.762

$\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696$

$\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784$

95% of the rates are expected to be betwen 0.696 and 0.784

$\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674$

$\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806$

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

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

So then:

$\mu_p = 0.74$

$\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219$

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

$\mu_p -\sigma_p = 0.74-0.0219=0.718$

$\mu_p +\sigma_p = 0.74+0.0219=0.762$

68% of the rates are expected to be betwen 0.718 and 0.762

$\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696$

$\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784$

95% of the rates are expected to be betwen 0.696 and 0.784

$\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674$

$\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806$

99.7% of the rates are expected to be betwen 0.674 and 0.806

5 0

3 years ago

A theater wants to build movable steps that they can use to go on and off the stage. They want the steps to have enough space in

nataly862011 [7]

Answer:

Step-by-step explanation:

Do you smart to party

8 0

4 years ago

Read 2 more answers

Verify identity list steps. Cot(t)(1-cos^2(t))=cos(t)sin(t)

storchak [24]

Remember: We have to work from either the LHS or the RHS.
(Left hand side or the Right hand side)

You should already know this:

$\huge{Cot(t) = \frac{1}{tan(t)} = \frac{1}{\frac{sin(t)}{cos(t)}} = 1\div \frac{sin(t)}{cos(t)} = 1\times \frac{cos(t)}{sin(t)}=\boxed{\frac{cos(t)}{sin(t)}}$

You should also know this:

$sin^2(t) + cos^2(t) = 1\\\\\boxed{sin^2(t)} = 1 - cos^2(t)$

So plugging in both of those into our identity, we get:

$\frac{cos(t)}{sin(t)}\cdot sin^2(t) = cos(t)\cdot sin(t)$

Simplify the denominator on the LHS (Left Hand Side)

We get:

$cos(t) \cdot sin(t) = cos(t) \cdot sin(t)$

LHS = RHS

Therefore, identity is verified.

4 0

3 years ago