A number was multiplied by 6 then increased by 3 and the result was 33

What is the side length a in the triangle below?

grigory [225]

Answer:

a=5

Step-by-step explanation:

a^2+b^2=c^2

c^2-b^2=a^2

13^2-12^2=a^2

169-144=a^2

25=a^2

5=a

8 0

3 years ago

If people blink about 13 times per minute, how many times do they blink in 7 hours?

SCORPION-xisa [38]

Answer:

Step-by-step explanation:

1 hour=60 min

7 hours=60×7=420 min

number of blink=420×13=5460

4 0

3 years ago

Hey!! I need this answer fast, no cap allowed!!

KIM [24]

1

(5(0)+1/2)2

(1/2)2

2/2

1

8 0

3 years ago

What is the value of RT or x

Anna007 [38]

Write and solve an equation of ratios:

x 9

---- = ----

16 x

Cross mult., we get x^2 = (16)(9)

Taking the sqrt of both sides, we get x = plus or minus (4)(3), or

x= plus or minus 12

Since x is measure of distance here, omit the -12 result. x = +12 units

7 0

3 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.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

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

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

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 $-0.1412 \leq p_A -p_B \leq -0.0318$

7 0

3 years ago