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3 years ago

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In a basketball game, the bulldogs make a total of 21 shots. Some of the shots are 2-pt shots while others are 3-pt shots. The b

ulldogs score a total of 50 points. How many 2-point and 3-point shots did they make?

1 answer:

VARVARA [1.3K]3 years ago

6 0

Answer:

It is 8 3 pointers and 13 2 pointers

Step-by-step explanation:

8 x 3 = 24

13 x 2 = 26

24 + 26 = 50

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Find all of the eigenvalues λ of the matrix A. (Hint: Use the method of Example 4.5 of finding the solutions to the equation 0 =

Svetradugi [14.3K]

Answer:

$\lambda=8,\ \lambda=-5$

Step-by-step explanation:

<u>Eigenvalues of a Matrix</u>

Given a matrix A, the eigenvalues of A, called $\lambda$ are scalars who comply with the relation:

$det(A-\lambda I)=0$

Where I is the identity matrix

$I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

The matrix is given as

$A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]$

Set up the equation to solve

$det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0$

Expanding the determinant

$det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0$

$(3-\lambda)(-\lambda)-40=0$

Operating Rearranging

$\lambda^2-3\lambda-40=0$

Factoring

$(\lambda-8)(\lambda+5)=0$

Solving, we have the eigenvalues

$\boxed{\lambda=8,\ \lambda=-5}$

8 0

3 years ago

Find the area of shaded part .

Firlakuza [10]

Answer:

Area of big rectangle= 5m× 3m

= 15m²

Area of small rectangle=1m× 1m

=1m²

Area of shaded part= Area of big rectangle- Area of a small rectangle

= 15m²-1m²

= 14m²

Step-by-step explanation: Hope it helps

5 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

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3 years ago

Anyone wanna be friends

boyakko [2]

Answer:

Yes :)

Step-by-step explanation:

7 0

3 years ago

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Identify the slope and y-intercept of a graph of the equation y=1/2x-5

ehidna [41]

Answer:

Slope: 1/2

y-intercept: -5

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3 years ago