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Ivenika [448]
3 years ago
6

Tommy thought of a number. He multiplied the number 4, then added six. The result was fifty. What was Tommy's starting number?

Mathematics
1 answer:
anygoal [31]3 years ago
5 0

Answer:

Was number 11

Step-by-step explanation:

50 - 6 = 44

44 dividid 4 is 11

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Given: Triangles ABC and DBC are isosceles, m∠BDC = 30°, and m∠ABD = 155°. Find m∠ABC, m∠BAC, and m∠DBC.
just olya [345]
∠BDC=30, so ∠CBD + ∠BCD=180-30=150
ΔDBC is isosceles, so ∠CBD=∠BCD=half of 150=75
∠ABC=∠ABD-∠CBD=155-75=80
ΔABC is isosceles, so ∠ABC=∠ACB=80
∠BAC=180-80-80=20
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4 years ago
The estimate regression equation for a model involving two independent variables and 10 observations follows.
True [87]

Answer:

a) b_1 = 0.5906\\b_2 = 0.4980

b) y = 289.815

Step-by-step explanation:

We are given the following information:

The estimate regression equation for a model involving two independent variables and 10 observations follows:

y = 29.1270 + 0.5906x_1 + 0.4980x_2

where x_1, x_2 are the two independent variable and y is the dependent variable.

a) The regression line is of the form:

y = b_0 + b_1x_1 + b_2x_2\\\text{where } b_0 \text{ is the y-intercept and }b_1, b_2\text{ are the coefficients of} x_1, x_2\text{ respectively}

Comparing, we get,

b_1 = 0.5906\\b_2 = 0.4980

These are the coefficients of x_1, x_2 respectively, and helps us to know the the significance of the independent variable in predicting the independent variable.

b) We have to estimate y when

x_1 = 180 \text{ and } x_2 = 310\\y = 29.1270 + 0.5906x_1 + 0.4980x_2\\\text{Putting values}\\y = 29.1270 + 0.5906(180) + 0.4980(320)\\y = 289.815

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4 years ago
Express 11.037 in words.
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Eleven and thirty seven hundredth
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Each stake has 12 onces
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3 years ago
At the beginning of year 1, Sam invests $700 at an annual compound interest
Aleksandr-060686 [28]

at the beginning of year 4, only 3 years have elapsed, the 4th year hasn't started yet, since it's at the beginning, so at the beginning of year 4 we can say only 4-1 years have elapsed.

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$700\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=\textit{elapsed years}\dotfill &3 \end{cases}

A=700\left( 1 + \frac{0.05}{1} \right)^{1\cdot 3}\implies A = 700(1+0.05)^3\implies A(4)=700(1+0.05)^{4-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A(n)=700(1+0.05)^{n-1}~\hfill

6 0
3 years ago
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