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3 years ago

13

PLEASE HELP! a rectangular picture frame has a perimeter of 58 inches.the height of the frame is 16 inches.what is the width of

the frame? tysm and pls show work

2 answers:

kipiarov [429]3 years ago

7 0

16+16= 32
58-32=26
26÷2= 13
so the width is 13 inches
13+13+16+16=
26 + 32
58

Vesnalui [34]3 years ago

3 0

The width is 13 cubic inches

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A ball is thrown into the air with an upward velocity of 36 ft/s. It’s height h in feet after t seconds is given by the function

777dan777 [17]

Answer:

Time taken for the ball to hit the ground back = 3.08 s

Step-by-step explanation:

h(t)= -16t² + 48t + 4

when will rhe object come back to hit rhe ground?

When the ball is at the level.of the ground, h(t) = 0.

0 = -16t² + 48t + 4

-16t² + 48t + 4 = 0

Solving the quadratic equation

t = 3.08 s or t = -0.08 s

Since the time cannot be negative,

Time taken for the ball to hit the ground back = 3.08 s

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Step-by-step explanation:

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3 years ago

The base of a rectangle is 15cm the perimeter is at most 94 cm. Write and solve an inequality to find the possible heights of th

tino4ka555 [31]

15 x 6 = 90 + 4 = 94 that's what I would do

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3 years ago

A computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) <u>Two computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

P(X = 2) = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) <u>Eight computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

(c) <u>Here, also 8 computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

<em>LET X = Number of computers classified as cutting dash edge</em>

So, it means X ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

= $1 - [1 \times 1 \times 0.94^{8}]$

= 1 - $0.94^{8}$ = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

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2 years ago

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Salsk061 [2.6K]

Answer:

X = 5

Step-by-step explanation:

28 = 7x + 3 - 2x

28 = 5x + 3

28 - 3 = 5x + 3 - 3

25 = 5x

25/5 = 5x/5

X = 5

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2 years ago

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Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7.

Alecsey [184]

<u><em>Answer:</em></u>

y^2 = 28x

<em><u>Step-by-step explanation:</u></em>

Since the directrix is horizontal, use the equation of a parabola that opens left or right.

(y−k)^2 = 4p(x−h)

Find the vertex.

(0,0)

Find the distance from the focus to the vertex.

p = 7

Substitute in the known values for the variables into the equation

(y−k)^2 = 4p(x−h).

(y−0)^2 = 4(7)(x−0)

Simplify.

<em>y^2 = 28x</em>

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3 years ago

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