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2 years ago

Find the measure of arc DB in Pplease hurry if you can

Mathematics

1 answer:

jeyben [28]2 years ago

8 0

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \:m\overbrace{DB}= 90 \degree$

____________________________________

$\large \tt Solution \: :$

$\qquad \tt \rightarrow \: m \overbrace{DB} + m \overbrace{TB} = 180 \degree$

[ linear pair ]

$\qquad \tt \rightarrow \: m \overbrace{DB} + 90 \degree = 180 \degree$

$\qquad \tt \rightarrow \: m \overbrace{DB} = 180 \degree - 90 \degree$

$\qquad \tt \rightarrow \: m \overbrace{DB} = 90 \degree$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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The sales tax in your city is 4.4% and a item costs 3$ before tax. How much tax would you pay

jolli1 [7]

Answer:

0.13 tax

Step-by-step explanation:

0.044*3=0.132

4 0

3 years ago

How many grams of a 15 percent alcohol solution should be mixed with 40 grams of a 30 percent solution to obtain a

Akimi4 [234]

The number of grams is 20 grams

Let x represent how many grams of a 15 percent alcohol solution

15%x+30%.40/x+40

=25%

0.15x+0.4.30=0.25 (x+40)

0.15x+12=0.25x+10

0.15x-0.25x=10+12

-0.1x=-2

x=-2/-0.1

x=20

Inconclusion The number of grams is 20 grams

Learn more here:

brainly.com/question/24750329

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3 years ago

Mrs.Chang spent $44 for a movie ticket and a snack for each of her 4 children. Each snack cost $3.50. Each movie ticket cost the

jeka57 [31]

Each snack cost 3.50 and she bought four, one for each child. The child who wrote it only took one snack into account.
The equation should be 14+4x=44
14 being all four snacks

8 0

3 years ago

A study of the number of business lunches that executives in the insurance and banking industries claim as deductible expenses p

gizmo_the_mogwai [7]

Answer:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

$p_v =2*P(t_{(88)}>2.604)=0.0108$

So the p value is a very low value and using any significance level for given $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Step-by-step explanation:

Data given and notation

$\bar X_{1}=9.1$ represent the mean for the sample 1

$\bar X_{2}=8$ represent the mean for the sample 2

$s_{1}=1.9$ represent the sample standard deviation for the sample 1

$s_{2}=2.1$ represent the sample standard deviation for the sample 2

$n_{1}=40$ sample size for the group 1

$n_{2}=50$ sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

Statistical decision

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

$df=n_{1}+n_{2}-2=40+50-2=88$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(88)}>2.604)=0.0108$

3 0

3 years ago

Please answer the following question

Arte-miy333 [17]

Yeah there all right but I have a little hinch for the last and first one

6 0

4 years ago

Find the measure of arc DB in Pplease hurry if you can​

Find the measure of arc DB in Pplease hurry if you can