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mestny [16]
3 years ago
11

I NEED HELP FAST ON THIS PROBLEM!!!!! Idk what the answer is!!

Mathematics
1 answer:
djyliett [7]3 years ago
4 0
Angle 1 = (140-40)/2

Angle 1 = 100/2
Angle 1 = 50 
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X 6.4.5-T
butalik [34]

Same strategy as before: transform <em>X</em> ∼ Normal(76.0, 12.5) to <em>Z</em> ∼ Normal(0, 1) via

<em>Z</em> = (<em>X</em> - <em>µ</em>) / <em>σ</em>   ↔ <em>X</em> = <em>µ</em> + <em>σ</em> <em>Z</em>

where <em>µ</em> is the mean and <em>σ</em> is the standard deviation of <em>X</em>.

P(<em>X</em> < 79) = P((<em>X</em> - 76.0) / 12.5 < (79 - 76.0) / 12.5)

… = P(<em>Z</em> < 0.24)

… ≈ 0.5948

8 0
3 years ago
Please help me I need a good grade I really need this or ,y mom will spank me please help me ¡!!!
AURORKA [14]

Answer:

A D and E

Step-by-step explanation:

3 0
3 years ago
Can someone please explain how to solve such a mean and sneaky problem? Please?
Lelu [443]

Answer:

all real numbers

Step-by-step explanation:

simplify each side of the inequality

26 + 6b -- already fully simplified

2(3b + 4) → 6b + 8

26 + 6b ≥ 6b + 8

subtract 6b from both sides

26 ≥ 8

since this expression is true (26 is greater than 8), then the solution is all real numbers. any value could take the place of b because in the end, it will be canceled out by itself, resulting in 26 ≥ 8

4 0
3 years ago
A grocery store manager uses 1/2 crate of apples for every 3/4 crate of oranges in a fruit display. How many crates of will she
mariarad [96]
15/4

3 3/4


Mark brainliest please
3 0
3 years ago
Read 2 more answers
Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30
Dafna11 [192]
Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

\bf u=&#10;\begin{cases}&#10;x=7cos(330^o)\\&#10;\qquad 7\cdot \frac{\sqrt{3}}{2}\\&#10;\qquad \frac{7\sqrt{3}}{2}\\&#10;y=7sin(330^o)\\&#10;\qquad 7\cdot -\frac{1}{2}\\&#10;\qquad -\frac{7}{2}&#10;\end{cases}\qquad \qquad v=&#10;\begin{cases}&#10;x=8cos(30^o)\\&#10;\qquad 8\cdot \frac{\sqrt{3}}{2}\\&#10;\qquad \frac{8\sqrt{3}}{2}\\&#10;y=8sin(30^o)\\&#10;\qquad 8\cdot \frac{1}{2}\\&#10;\qquad 4&#10;\end{cases}

\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)&#10;\\\\\\&#10;\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~  \stackrel{b}{\frac{1}{2}}\right)\\\\&#10;-------------------------------

\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}&#10;\\\\\\&#10;\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o
8 0
3 years ago
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