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3 years ago

Help me please !!!!

Mathematics

1 answer:

fiasKO [112]3 years ago

6 0

Answer:

BC = 10

Step-by-step explanation:

BC = BD - CD

BC = 14 - 4

BC = 10

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Combine the like terms to create an equivalent expression: \large{5n+6+(-7n)}5n+6+(−7n)

forsale [732]

Answer:−2n+6,−2n+6

Step-by-step explanation:

5n+6−7n,5n+6−7n

((5n−7n)+6),((5n−7n)+6)

−2n+6,5n−7n+6

−2n+6,((5n−7n)+6)

−2n+6,−2n+6

7 0

3 years ago

5x - 4 > 12 AND 12x +5≤-4

LenaWriter [7]

Answer:

These two would not overlap each other when graphed so there is no solution.

6 0

3 years ago

Anyone solve this maths questions from integers

gogolik [260]

Step-by-step explanation:

SEE THE IMAGE FOR SOLUTION 

4 0

3 years ago

Solve each equation 1. r - 38 = 9 2. h + 17 = 40 3. n + 75 = 155 4. q - 17 = 18

taurus [48]

Answer:

1. r= 47

2.h=23

3.n=80

4.q=35

Step-by-step explanation:

r=9+38

h=40-17

n=155-75

q=18+17

6 0

3 years ago

The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they

Airida [17]

Answer:

a) Null hypothesis: $\mu \geq 14$

Alternative hypothesis: $\mu < 14$

H0: μ -greater than or equal to 14

Ha: μ - less than 14

b) $p_v =P(t_{(49)}$

c) For this case since the p value is higher than the significance level given of 0.05 the answer would be NO

We can't conclude that the actual mean waiting time is significantly less than the claim of 14 minutes made by the taxpayer advocate

Step-by-step explanation:

Data given

$\bar X=12$ represent the sample mean

$\sigma=10$ represent the population standard deviation

$n=50$ sample size

$\mu_o =14$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 14 minute, the system of hypothesis would be:

Null hypothesis: $\mu \geq 14$

Alternative hypothesis: $\mu < 14$

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Calculate the statistic

If we replace we got:

$t=\frac{12-14}{\frac{10}{\sqrt{50}}}=-1.414$

Part b: P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=50-1=49$

Since is a one sided lower test the p value would be:

$p_v =P(t_{(49)}$

Part c

For this case since the p value is higher than the significance level given of 0.05 the answer would be NO

We can't conclude that the actual mean waiting time is significantly less than the claim of 14 minutes made by the taxpayer advocate

8 0

3 years ago