Can someone help me with this problem! I WILL GIVE A BRAINIEST!!! I need it asap PLEASE.

Julia is going to the store to buy candies. Small candies cost $4 and extra-large candies cost $12.She needs to purchase at leas

BartSMP [9]

Answer:

Small candies $=9$

Extra large candies $=12$

Step-by-step explanation:

Let small candies $=x$

Extra large candies $=y$

the number of candies is at least $20$ .

$x+y\geq20$

Cost of $1$ small candy $=\$4$

Cost of $1$ extra large candy $=\$12$

but she has only $\$180$ to spend

$4x+12y\leq180$

Solve for

$x+y=20.......(1)\\4x+12y=180.....(2)\\eqn(2)-eqn(1)\times4\\8y=100\\y=\frac{100}{8} \\y=\frac{25}{8} \\from\ eqn(1)\\x+\frac{25}{2}=20\\ x=20-\frac{25}{2} \\x=\frac{15}{2}$

Since number of candies should be integer.

let $x=7,y=13$

total spend $4\times7+12\times13=184$ which is more than $\$180$ , so this combination is not possible.

$let\ x=8,y=12\\8\times4+12\times12=176$

She has $\$4$ more so she can buy $1$ more small candy.

Hence small candy $=9$

extra large candy $=12$

4 0

2 years ago

Question 2. Do I divide 9.76 by 8 or the opposite?

schepotkina [342]

It could go either way

7 0

3 years ago

Read 2 more answers

The assembly line that produces an electronic component of a missile system has historically resulted in a 2% defectiverate. a r

sattari [20]

<span>Defective rate can be expected to keep an eye on a Poisson distribution. Mean is equal to 800(0.02) = 16, Variance is 16, and so standard deviation is 4.
X = 800(0.04) = 32, Using normal approximation of the Poisson distribution Z1 = (32-16)/4 = 4.
P(greater than 4%) = P(Z>4) = 1 – 0.999968 = 0.000032, which implies that having such a defective rate is extremely unlikely.</span>

<span>If the defective rate in the random sample is 4 percent then it is very likely that the assembly line produces more than 2% defective rate now.</span>

3 0

3 years ago

Write the third missing congruent statement that will allow us to prove the triangles congruent by SAS.

musickatia [10]

Answer:

LOOK AT YOUR ANSWER IN PICTURE

8 0

3 years ago

Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a screening

Finger [1]

Answer:

A) Alternative hypothesis; Ha: p_o > ⅔

B) Null hypothesis;

H0: p_o = ⅔

C) There is sufficient evidence to support the claim that the community screening programme was effective

D) p-value = 0

Its less than the significance value and so we will reject the null hypothesis and conclude that the community screening programme was effective

Step-by-step explanation:

A) We are told that ⅓ of those diagnosed eventually die of the disease. Thus, ⅔ is the fraction that will survive the disease.

Thus, the null hypothesis is;

H0: p_o = ⅔

We are told that a screening programme was started to increase the survival rate. Thus,the alternative hypothesis is;

Ha: p_o > ⅔

B) null hypothesis is;

H0: p_o = ⅔

C) 164 out of the 200 women selected survived the disease.

Thus sample proportion is;

p^ = 164/200

p^ = 0.82

For the z-score value, we will use the formula;

z = (p^ - p_o)/√((p_o(1 - p_o)/n)

Now, p_o = ⅔ = 0.67

Thus;

z = (0.82 - 0.67)/√((0.67(1 - 0.67)/200)

z = 4.51

From online p-value from z-score calculator, we have p-value ≈ 0

This is less than the significant level of 0.05 and therefore we will reject the null hypothesis and conclude that the community screening program was effective.

D) p-value ≈ 0

Reject the null hypothesis

4 0

2 years ago