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3 years ago

Considere the diagram below determine DHG

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

6 0

9514 1404 393

Answer:

A) ∠DHG = 96°

B) arc EF = 134°

Step-by-step explanation:

<u>Part A</u>:

As you have shown in your work, angle DHG is supplementary to given angle FHG:

∠DHG = 180° -∠FHG

∠DHG = 180° -84°

∠DHG = 96°

<u>Part B</u>:

∠DHG is half the sum of arcs EF and DG.

96° = (1/2)(arc EF + 58°) . . . . . fill in known values

192° -58° = arc EF . . . . . . . . . . multiply by 2, subtract 58°

arc EF = 134°

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The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 i

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Null hypothesis: $\mu \geq 250$

Alternative hypothesis: $\mu < 250$

$p_v =P(t_{15}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=510$ represent the mean for the sample

$s=50$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =530$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

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

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$