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3 years ago

Find the measure of each angle. m

Mathematics

1 answer:

sleet_krkn [62]3 years ago

4 0

Explain more please in order to understand ? Is there a picture

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Rose spent 124 on 10 cars & 4 cats what is the cost for 1 car & 1 cat.

Rudiy27

Answer:

$6 per cat, $10 per car

Step-by-step explanation:

I guess and checked my work.

4 0

3 years ago

Joey and Armando live on the same st as the park. The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando'

AleksandrR [38]

Let's start by assuming Armando's house is between Joey's and the park.

Let $x$ be the distance Joey walked to Armando's house.

The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando's home. Then Joey and Armando walk 3/5 mile to the park.

 $\dfrac{9}{10} = x + \dfrac{3}{5}$

$x = \dfrac{9}{10} - \dfrac{3}{5} = \dfrac{9}{10} -\dfrac{6}{10} = \dfrac{3}{10}$

That's probably the answer they're looking for. But what if the park is between Joey and Armando's houses or Joey is between the park and Armando? (The latter isn't really possible with the given distances.)

Let $a, b, c$ be the distances between three collinear points like we have here. Our equation is really a few equations in one, something like

$\pm a \pm b = \pm c$

Let's get rid of the plus/minuses. Squaring,

$a^2 + b^2\pm 2ab = c^2$

$\pm 2ab = c^2-a^2-b^2$

$4a^2b^2 = (c^2-a^2-b^2)^2$

For us, that's a quadratic equation for $c^2$

$4(9/10)^2(3/5)^2= (c^2-(9/10)^2 - (3/5)^2)^2$

I'll skip right to the solutions,

$c^2=\dfrac{9}{100} \textrm{ or } c^2=\dfrac{9}{4}$

$c=\dfrac{3}{10} \textrm{ or } c=\dfrac{3}{2}$

We could have gotten the 3/2 just by adding 9/10+3/5 but this was more fun.

6 0

4 years ago

1x + 2 - 3 = 8 what is x?

valina [46]

Answer:

x= Multiplication

Step-by-step explanation:

1 x +2 -3 =8

8 0

3 years ago

Read 2 more answers

2. Which expression is equivalent to 3x(-4x² + 5x - 8) - 6(x² - 5x + 7)?

adoni [48]

It is answer choice B. Using the distributive property, -12x^3+15x^2-24x-6x^2+30x-42. Then combine like terms.

5 0

3 years ago

Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask

Kay [80]

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

$P(x=r)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

$P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}$

$=56(0.4)^5 (0.6)^3$

$=0.12386304$

(b) the probability of the number that say they would feel secure is more than five

$P(X>5) = P(X=6)+ P(X=7) + P(X=8)$

$=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}$

$=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8$

$=0.04980736$

(c) the probability of the number that say they would feel secure is at most five

$P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}$

$=0.6^8+8(0.4)(0.6)^7+28(0.4)^2(0.6)^6+56(0.4)^3(0.6)^5+70(0.4)^4(0.6)^4+56(0.4)^5(0.6)^3$

$=0.95019264$

8 0

3 years ago