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2 years ago

I need help ASAP snsnnsbsbsbs

Mathematics

2 answers:

harkovskaia [24]2 years ago

4 0

The value is 50 hope it helped

anastassius [24]2 years ago

3 0

Answer:

The value of a is 50°.

Step-by-step explanation:

In an isosceles triangle, 2 sides have the same lengths and same angles so we can assume that the other side of the triangle is 65°.

Given that the total angle of a triangle is 180°, so have to subtract it :

$a = 180 - 65 - 65$

$a = 180 - 130$

$a = 50$

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Amiraneli [1.4K]

Answer:

Step-by-step explanation:

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2 years ago

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 18 m

Colt1911 [192]

Answer:

99.7% of customers have to wait between 8 minutes to 30 minutes for their food.

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 18 minutes

Standard Deviation, σ = 4 minutes

We are given that the distribution of amount of time is a bell shaped distribution that is a normal distribution.

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, 99.7% of the customers have to wait:

$\mu -3\sigma = 18-3(4) = 6\\\mu +3\sigma = 18+3(4) = 30$

Thus, 99.7% of customers have to wait between 8 minutes to 30 minutes for their food.

3 0

3 years ago

8xf(0)+4xg(-8)= solve this

Virty [35]

The Answer is: -32xg

7 0

3 years ago

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

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3 years ago

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Gekata [30.6K]

Answer:

42% i hope its right if not sorry

Step-by-step explanation:

4 0

2 years ago