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3 years ago

13

Sue needs 2 cups of flour for a batch of cookies. how many complete batches can she make with 10 cups of flour? (1 point) 5 batc

hes 3 batches 2 batches 4 batch

1 answer:

Arlecino [84]3 years ago

7 0

Cupshas/cupsneededperecepie=recepies possible

has 10 cups
needs 2 cups per recepie

10/2=5

5 batches

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Find each sum (-4y+3) + (11y-5)

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2 years ago

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
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$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

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2 years ago

Is shipping box is a shape of a cube. Each side measures 3c^6d^2 inches. Express the volume of a cube has a monomial.

liubo4ka [24]

Length of each side of cube is = $3c^6d^2$

SO the volume of cube = $(Side)^3$
= $(3c^6d^2)^3$
= $27 * (c^6)^3(d^2)^3$
= $27c^{18}d^6$

SO the volume of cube is $27c^{18}d^6$

9 0

3 years ago

Read 2 more answers

In a random sample of 8 people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.4 minutes. A 90

konstantin123 [22]

Answer:

Margin of Error = 5.4088 ;

Confidence interval = (30.1 ; 40.9)

Interval estimate are almost the same

Step-by-step explanation:

Given that :

Population standard deviation, σ = 9.3

Sample size, n = 8

Xbar = 35.5

Confidence level = 90%

The confidence interval:

Xbar ± Margin of error

Margin of Error = Zcritical * σ/sqrt(n)

Zcritical at 90% = 1.645

Margin of Error = 1.645 * 9.3/sqrt(8) = 5.4088

Confidence interval :

Xbar ± Margin of error

35.5 ± 5.4088

Lower boundary = (35.5 - 5.4088) = 30.0912 = 30.1

Upper boundary = (35.5 + 5.4088) = 40.9088 = 40.9

(30.1 ; 40.9)

T distribution =. (30.5 ; 40.5)

Normal distribution = (30.1, 40.9)

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2 years ago