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3 years ago

First picture is just information second picture please answer I will mark brainliest

Mathematics

2 answers:

Alecsey [184]3 years ago

8 0

Answer:

2329

Step-by-step explanation:

ziro4ka [17]3 years ago

6 0

Answer:

2329 cups are left at the end of the week

Step-by-step explanation:

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What is the answer to this?

nekit [7.7K]

Answer:

Step-by-step explanation:

Y= 1x^2 + 3x + 9

Y= Ax^2 + Bx + C

To determine which way a parabola opens just look at the coefficient (the number) of the x^2 - Always use the coefficient that is beside the x^2

So, if the coefficient A is positive the parabola is opening upward

if the coefficient A is negative the parabola is opening downward

7 0

2 years ago

Read 2 more answers

Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

3 years ago

The largest orange in a bag has the circumference of 9 5/8 inches. The smallest orange has a circumference

Marysya12 [62]

Answer:

1.81 inches

Step-by-step explanation:

Given that,

The circumference of the largest orange in a bag is $9\dfrac{5}{8}\ \text{inches}$ and that of smallest orange is $7\dfrac{13}{16}\ \text{inches}$ .

Difference = largest orange - smallest orange

= $9\dfrac{5}{8}-7\dfrac{13}{16}\\\\=\dfrac{77}{8}-\dfrac{125}{16}\\\\=\dfrac{29}{16}\\\\=1.81\ \text{inches}$

So, difference of the circumferences of the smallest orange and the largest orange is 1.81 inches

4 0

3 years ago

Tara is making dinner for 9 people. She mixes 1 1/4 kilograms of rice with 500grams of vegetables and 410 grams of meat. If all

kotegsom [21]

Answer:

240 grams

Step-by-step explanation:

Tara uses:

$1 \dfrac{1}{4}\ kg=1 \ kg\ 250 \ g=1250\ g$ of rice
500 g of vegatables
410 g of meat

In total, Tara uses

$1250+500+410=2160\ g$ of ingredients.

She is making dinner for 9 people, so the weight of each serving is

$2160:9=240\ g$

3 0

3 years ago

Solve the problem. Round as appropriate.

Monica [59]

Answer:

i need the points

Step-by-step explanation:

8 0

3 years ago