All categories

3 years ago

-5(11-3t) = -6t - 4 what does t equal?

1 answer:

6 0

-5(11-3t) = -6t -4

distribute -5 to the parentheses

-55+15t = -6t -4

isolate t; add -6t from both sides

-55+21t = -4

add 55 from both sides

21t = 51

divide t

t = 2.4 (rounded)

You might be interested in

Will someone help me with this one I'm not good at simplifying

arlik [135]

A is 0.0028
and B is 0.054

8 0

3 years ago

Read 2 more answers

What is 22.5x3.32 and Show your Work

Nataliya [291]

Answer:

74.7

Step-by-step explanation:

Just use muliplication

8 0

4 years ago

The expression 2+√3 over 2-√3 is equivalent to

Dimas [21]

$a : 11 \sqrt{3}$

4 0

3 years ago

A college entrance exam company determined that a score of 24 on the mathematics portion of the exam suggests that a student is

seraphim [82]

Answer:

a) The Z -value 2.397 < 2.576 at 99% or 0.01% level of significance

Null hypothesis is accepted at 0.01% level of significance

They score is above 24 on the math portion of the exam

b)

Null Hypothesis: There is no significance difference between the college level mathematics and math courses in high school

H₀: μ = 24

Alternative Hypothesis: H₁: μ ≠ 24

Step-by-step explanation:

Step(i):-

Given random sample 'n' = 250

Given data sample mean x⁻ = 24.5

Standard deviation = 3.3

Null Hypothesis: There is no significance difference between the college level mathematics and math courses in high school

H₀: μ = 24

Alternative Hypothesis: H₁: μ ≠ 24

test statistic

$Z = \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{24.5 - 24}{\frac{3.3}{\sqrt{250} } } = \frac{0.5}{0.2087} = 2.397$

a) 99% or 0.01% level of significance

Level of significance ∝ = 0.01

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.005} =2.576$

The Z -value 2.397 < 2.576 at 99% or 0.01% level of significance

Null hypothesis is accepted at 0.01% level of significance

They score is above 24 on the math portion of the exam

b) 95% or 0.05% level of significance

Level of significance ∝ = 0.05

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96$

The Z -value 2.397 > 1.96 at 95% or 0.05% level of significance

Null hypothesis is Rejected at 0.05% level of significance

They score is below 24 on the math portion of the exam

6 0

3 years ago

In a survey of US adults, twenty percent say they never exercise. This is the highest level seen in five years.

Andrews [41]

Answer:

$0.20 - 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.175$

$0.20 + 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.225$

The 95% confidence interval would be given by (0.175;0.225)

Step-by-step explanation:

Assuming this question: In a survey of 1000 US adults, twenty percent say they never exercise. This is the highest level seen in five years.1 Find a 95% confidence interval for the proportion of US adults who say they never exercise. Round your answers to three decimal places.

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".

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

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

If we replace the values obtained we got:

$0.20 - 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.175$

$0.20 + 1.96\sqrt{\frac{0.2(1-0.2)}{1000}}=0.225$

The 95% confidence interval would be given by (0.175;0.225)

6 0

3 years ago