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malfutka [58]
3 years ago
9

How do you solve the height?​

Mathematics
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

If you know the base and area of the triangle, you can divide the base by 2, then divide that by the area to find the height. To find the height of an equilateral triangle, use the Pythagorean Theorem, a^2 + b^2 = c^2.

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If one lap takes me 11 minutes and 10 seconds how long will it take me to do 10 laps
weqwewe [10]

Answer:

111 minutes and 40 sec

Step-by-step explanation:

11x10=110 minutes

10 sec x 10 = 100 seconds = 1 minute 40 sec

110 minutes+1 minute+40 seconds= 111 minutes and 40 seconds

3 0
2 years ago
Which system of the equation is represents by the graph?<br><br><br>​
pashok25 [27]
Sorry just needed points to use this app
3 0
2 years ago
A bank features a savings account that has an annual percentage rate of 3.5% with interest compounded quarterly. Vanessa deposit
vazorg [7]

Answer:

APY = 0.04 or 4%

Step-by-step explanation:

Given the annual percentage rate of 3.5% that is compounded quarterly, and a principal of $6,500:

We can use the following formula to solve for the annual percentage yield (APY):

APY =  (1 + \frac{r}{n})^n - 1

where <em>r</em> = interest rate = 3.5% or 0.035

<em>           n</em> = number of compounding periods per year = 4

We can plug in the values into the equation:

APY =  (1 + \frac{r}{n})^n - 1

APY =  (1 + \frac{.035}{4})^4 - 1

APY =  (1 + 0.00875)^4 - 1

APY =  (1.00875)^4 - 1

APY = 1.03546 - 1  

APY = 0.04 or 4%        

8 0
2 years ago
How do I find a2 and a3 for the following geometric sequence? 54, a2, a3, 128
vlada-n [284]
The formula for the nth term of a geometric sequence:
a_n=a_1 \times r^{n-1}
a₁ - the first term, r - the common ratio

54, a_2, a_3, 128 \\ \\&#10;a_1=54 \\&#10;a_4=128 \\ \\&#10;a_n=a_1 \times r^{n-1} \\&#10;a_4=a_1 \times r^3 \\&#10;128=54 \times r^3 \\&#10;\frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\&#10;\frac{64}{27}=r^3 \\&#10;\sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\&#10;\frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\&#10;r=\frac{4}{3}

a_2=a_1 \times r= 54 \times \frac{4}{3}=18 \times 4=72 \\&#10;a_3=a_2 \times r=72 \times \frac{4}{3}=24 \times 4=96 \\ \\&#10;\boxed{a_2=72, a_3=96}
7 0
3 years ago
Read 2 more answers
Suppose you can somehow choose two people at random who took the SAT in 2014. A reminder that scores were Normally distributed w
Sindrei [870]

Answer:

22.29% probability that both of them scored above a 1520

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 1497, \sigma = 322

The first step to solve the question is find the probability that a student has of scoring above 1520, which is 1 subtracted by the pvalue of Z when X = 1520.

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{1520 - 1497}{322}

Z = 0.07

Z = 0.07 has a pvalue of 0.5279

1 - 0.5279 = 0.4721

Each students has a 0.4721 probability of scoring above 1520.

What is the probability that both of them scored above a 1520?

Each students has a 0.4721 probability of scoring above 1520. So

P = 0.4721*0.4721 = 0.2229

22.29% probability that both of them scored above a 1520

8 0
3 years ago
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